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EUCLID'S ELEMENTS OF GEOMETRY The Greek text of J.L. Heiberg (1883-1885) from Euclidis Elementa, edidit et Latine interpretatus est I.L. Heiberg, in aedibus B.G. Teubneri, 1883-1885 edited, and provided with a modern English translation, by Richard Fitz pat rick First edition - 2007 Revised and corrected - 2008 ISBN 978-0-6151-7984-1 Contents Introduction 4 Book 1 5 Book 2 49 Book 3 69 Book 4 109 Book 5 129 Book 6 155 Book 7 193 Book 8 227 Book 9 253 Book 10 281 Book 11 423 Book 12 471 Book 13 505 Greek-English Lexicon 539 Introduction Euclid's Elements is by far the most famous mathematical work of classical antiquity, and also has the distinction of being the world's oldest continuously used mathematical textbook. Little is known about the author, beyond the fact that he lived in Alexandria around 300 BCE. The main subjects of the work are geometry, proportion, and number theory. Most of the theorems appearing in the Elements were not discovered by Euclid himself, but were the work of earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus of Athens, and Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously discovered theorems: e.g., Theorem 48 in Book 1. The geometrical constructions employed in the Elements are restricted to those which can be achieved using a straight-rule and a compass. Furthermore, empirical proofs by means of measurement are strictly forbidden: i.e., any comparison of two magnitudes is restricted to saying that the magnitudes are either equal, or that one is greater than the other. The Elements consists of thirteen books. Book 1 outlines the fundamental propositions of plane geometry, includ- ing the three cases in which triangles are congruent, various theorems involving parallel lines, the theorem regarding the sum of the angles in a triangle, and the Pythagorean theorem. Book 2 is commonly said to deal with "geometric algebra", since most of the theorems contained within it have simple algebraic interpretations. Book 3 investigates circles and their properties, and includes theorems on tangents and inscribed angles. Book 4 is concerned with reg- ular polygons inscribed in, and circumscribed around, circles. Book 5 develops the arithmetic theory of proportion. Book 6 applies the theory of proportion to plane geometry, and contains theorems on similar figures. Book 7 deals with elementary number theory: e.g., prime numbers, greatest common denominators, etc. Book 8 is concerned with geometric series. Book 9 contains various applications of results in the previous two books, and includes theorems on the infinitude of prime numbers, as well as the sum of a geometric series. Book 10 attempts to classify incommen- surable {i.e., irrational) magnitudes using the so-called "method of exhaustion", an ancient precursor to integration. Book 11 deals with the fundamental propositions of three-dimensional geometry. Book 12 calculates the relative volumes of cones, pyramids, cylinders, and spheres using the method of exhaustion. Finally, Book 13 investigates the five so-called Platonic solids. This edition of Euclid's Elements presents the definitive Greek text — i.e., that edited by J.L. Heiberg (1883- 1885) — accompanied by a modern English translation, as well as a Greek-English lexicon. Neither the spurious books 14 and 15, nor the extensive scholia which have been added to the Elements over the centuries, are included. The aim of the translation is to make the mathematical argument as clear and unambiguous as possible, whilst still adhering closely to the meaning of the original Greek. Text within square parenthesis (in both Greek and English) indicates material identified by Heiberg as being later interpolations to the original text (some particularly obvious or unhelpful interpolations have been omitted altogether) . Text within round parenthesis (in English) indicates material which is implied, but not actually present, in the Greek text. My thanks to Mariusz Wodzicki (Berkeley) for typesetting advice, and to Sam Watson & Jonathan Fenno (U. Mississippi), and Gregory Wong (UCSD) for pointing out a number of errors in Book 1. 4 ELEMENTS BOOK 1 Fundamentals of Plane Geometry Involving Straight-Lines 5 ETOIXEIfiN a'. ELEMENTS BOOK 1 "Opoi. a'. Er^elov eoxiv, ou ^epoc; ou'dev. P'. rpa^rj 8s nfjxoc; a7tXocxe<;. y'. rpa^urj<; Be Ttepaxoc or]^eTa. 8'. EO'deToc ypo^iuf| eoxiv, ffiic, 15 taou tou; ecp' eauxrji; ar)fie[oi<; xelxai. e'. 'Eracpdveioc Be eoxiv, o urjxoc; xdi rcXdxoc; [iovov e/ei. <r'. Ti/iucpavdac; 8s Ttepaxa ypajjijiai. C- 'EtutteSoc; emcpdveid eoxiv, fjxic; ioou xaig ecp' eauxrjc; euT&eioac; xelxai. T)'. 'EtutccBoc; Be ytovia eaxiv f] ev sniTieScp Buo ypaji^Gv dnxo^ieviov dXXrjXwv xal u/] en' eui5eiac; xeijievtov npoc; dXXrjXac; xfiv ypa^cov x^ou;. "Oxav Be ai nspis/ouoai xr]v ywviav ypajjijidi eu-delai Goiv, eMuypaji^ioc; xaXeTxai f] ywvia. i'. "Oxav Be euiikTa in' eui&eTav oxa^eloa xdg ecpe^rjc; ywviac; Toac; dXXfjXaic; Ttoifj, 6p$r) exaxepa xfiv Tacov yioviCSv eoxi, xdi f) ecpeaxr)xuia eui9eTa xdi^exoc; xaXeTxai, ecp' fjv £(p£aXT)X£V. ia'. AjipXeTa ycovia eoxiv f] jiei^cjv op-drjc;. iP'. 'Ocelot 8s: f] eXdaocov op^rjc;. iy'. "Opoc; eoxiv, o xivoc; eoxi Ttepotc;. 18'. Sx^K a ^ OTl TO UIto Tlvo ? ^ xivwv opwv Ttepiexo^ievov. is'. KuxXoc; eoxi ox/jjia emiteBov Otto jiidc; ypajiurjc; TiEpiExojisvov [f] xaXsTxai nepicpepeia], npbc, fjv dcp' evoc; or^eiou xfiv evxoc; xou ax^axog xei^ieviov Ttdoai di npooTUTixouoai eu-delai [npog xrjv xoO xuxXou rcepicpepeiav] lacci dXXrjXaic; eiaiv. i=r'. Kevxpov 8s xou xuxXou xo or^eTov xaXeixai. Aidjiexpog 8s xou xuxXou eaxiv euaSe'id xic; Std xou xevxpou yjy^evr] xdi Tiepaxoujjievr) ecp' exdxepa xd jiepr) utto xrjc; xou xuxXou nepicpepeiac;, fjxic; xdi Si/a xeuvei xov xuxXov. it]'. 'HjiixuxXiov Be eoxi xo Ttepiex6|ievov ox^a Otto xe xrjc Biajjixpou xdi xrjg duoXa(jipavo|jievr]^ On' auxfjc; nepi- 9epeiag. xevxpov 8e xou f)^ixuxX[ou xo auxo, o xdi xou xuxXou eaxiv. it}'. S)(r]jiaxa eui9uypaji|jid eoxi xd utto eu-deifiv rce- pie/o^ieva, xpiicXeupa ^iev xd uno xpiwv, xexpdnXeupa Se xd utio xeoodpwv, noXunXeupa Se xd uno tiXciovcov f\ xeaadpwv eu-deifiv nepiexo^ieva. x'. TGv Be xpiTiXeupwv o/iqtidxwv iaoTiXeupov [ie\ xpiycovov eoxi xo xdg xpelg Taac; exov TiXeupdc;, ioooxeXec; 8e xo xdg 8uo [lovolc, Icolc, exov TtXeupdc;, oxaXrjvov 8e xo xdc; xpelg dviooug exov nXeupdc;. xa' "Exi 8e x«v xpiTiXeupwv oxr)|jidxwv 6pi9oycoviov [iev xpiycovov eoxi xo exov op'driv ycoviav, djipXuycjviov 8e xo exov djipXelav ywviav, o^uywviov 8e xo xdg xpelc; o^eiac; exov ywvia^. Definitions 1. A point is that of which there is no part. 2. And a line is a length without breadth. 3. And the extremities of a line are points. 4. A straight-line is (any) one which lies evenly with points on itself. 5. And a surface is that which has length and breadth only. 6. And the extremities of a surface are lines. 7. A plane surface is (any) one which lies evenly with the straight-lines on itself. 8. And a plane angle is the inclination of the lines to one another, when two lines in a plane meet one another, and are not lying in a straight-line. 9. And when the lines containing the angle are straight then the angle is called rectilinear. 10. And when a straight-line stood upon (another) straight-line makes adjacent angles (which are) equal to one another, each of the equal angles is a right-angle, and the former straight-line is called a perpendicular to that upon which it stands. 11. An obtuse angle is one greater than a right-angle. 12. And an acute angle (is) one less than a right-angle. 13. A boundary is that which is the extremity of some- thing. 14. A figure is that which is contained by some bound- ary or boundaries. 15. A circle is a plane figure contained by a single line [which is called a circumference], (such that) all of the straight-lines radiating towards [the circumference] from one point amongst those lying inside the figure are equal to one another. 16. And the point is called the center of the circle. 17. And a diameter of the circle is any straight-line, being drawn through the center, and terminated in each direction by the circumference of the circle. (And) any such (straight-line) also cuts the circle in half.^ 18. And a semi-circle is the figure contained by the diameter and the circumference cuts off by it. And the center of the semi-circle is the same (point) as (the center of) the circle. 19. Rectilinear figures are those (figures) contained by straight-lines: trilateral figures being those contained by three straight-lines, quadrilateral by four, and multi- lateral by more than four. 20. And of the trilateral figures: an equilateral trian- gle is that having three equal sides, an isosceles (triangle) that having only two equal sides, and a scalene (triangle) that having three unequal sides. 6 ETOIXEIfiN a'. ELEMENTS BOOK 1 xp'. Twv Se xexpaTiXeupwv axr^dxtov xexpdyovov (iev eoxiv, 8 EaoTtXeupov xe eaxi xal op-doywviov, exepo^xet; 8e, 8 op-doytoviov ^lev, oux looTtXeupov Se, po^poc; 8e, 8 iooTtXeupov jiev, oux opiDoycoviov 86, po^poeiBei; 8s xo xdc; dmevavxiov TiXeup&c xe xal ywvtac laac, dXXr|Xoa<; £X ov j ° ouxe taoTiXeupov eaxiv oOxs op'Ooycoviov xd 8e itapd xauxa xexpaitXeupa xpane^ia xaXeurdco. xy'. IIapdXXr)Xo[ rioiv eii/dsToci, atxivec; ev tw auxw eraTieBo ouoai xal expaXXo^tevai sic; aTteipov ecp' exdxepa xd [lepr] era ^tr]8exepa au^mrnxouoiv dXXr]Xai<;. 21. And further of the trilateral figures: a right-angled triangle is that having a right-angle, an obtuse-angled (triangle) that having an obtuse angle, and an acute- angled (triangle) that having three acute angles. 22. And of the quadrilateral figures: a square is that which is right-angled and equilateral, a rectangle that which is right-angled but not equilateral, a rhombus that which is equilateral but not right-angled, and a rhomboid that having opposite sides and angles equal to one an- other which is neither right-angled nor equilateral. And let quadrilateral figures besides these be called trapezia. 23. Parallel lines are straight-lines which, being in the same plane, and being produced to infinity in each direc- tion, meet with one another in neither (of these direc- tions) . t This should really be counted as a postulate, rather than as part of a definition. AiTTjfiorca. a'. 'Hixrjcrdco duo navxoc; ay]y.eiov era raw arjueTov eMelav ypa^rjv dyayelv. P'. Kal TC£7iepaa^£v/)v eMelav xaxd xo auve/ec; in eCWteiac; expaXeTv. y'. Kal Ttavxl xevxpcp xal Biaaxr^axi xuxXov ypdcpecrdai. 8'. Kal naooLc, xd<; 6pi9a<; ycovtac laac dXXr]Xai<; elvai. z. Kal edv sic, 8uo eO'deiat; eui^sTa s^TUTtxouaa xdc; evxoc xal Era xa auxd [ispf] ywviac; Suo opiJcov eXdaaovag raoirj, expaXXojievac; xag 86o eu-deiac; in aTteipov au^iraTTxeiv, ecp' a y.spi] eialv ai x«v Suo 6pi9Gv eXdaaovec;. Postulates 1. Let it have been postulated* to draw a straight-line from any point to any point. 2. And to produce a finite straight-line continuously in a straight-line. 3. And to draw a circle with any center and radius. 4. And that all right-angles are equal to one another. 5. And that if a straight-line falling across two (other) straight-lines makes internal angles on the same side (of itself whose sum is) less than two right-angles, then the two (other) straight-lines, being produced to infinity, meet on that side (of the original straight-line) that the (sum of the internal angles) is less than two right-angles (and do not meet on the other side).* t The Greek present perfect tense indicates a past action with present significance. Hence, the 3rd-person present perfect imperative 'HiT^cnSa) could be translated as "let it be postulated", in the sense "let it stand as postulated", but not "let the postulate be now brought forward". The literal translation "let it have been postulated" sounds awkward in English, but more accurately captures the meaning of the Greek, t This postulate effectively specifies that we are dealing with the geometry of flat, rather than curved, space. Koivat svvoiai. Common Notions a'. Td xw auxw laa xal dXXrjXou; eaxlv loa. P'. Kal edv laoic I'oa TCpoaxcdrj, xd oXa eaxlv laa. y'. Kal edv duo ia«v I'oa dcpaipsiS/j, xd xaxaXeiTto^ievd eaxiv I'oa. 8'. Kal xd ecpap[i6£ovxa iiC dXXf]Xa loa dXXrjXou; eaxiv. e'. Kal xo oXov xou [Lepouz [leiZov [eaxiv]. 1. Things equal to the same thing are also equal to one another. 2. And if equal things are added to equal things then the wholes are equal. 3. And if equal things are subtracted from equal things then the remainders are equal.* 4. And things coinciding with one another are equal to one another. 5. And the whole [is] greater than the part. t As an obvious extension of C.N.s 2 & 3 — if equal things are added or subtracted from the two sides of an inequality then the inequality remains 7 ETOIXEIfiN a'. ELEMENTS BOOK 1 an inequality of the same type. a'. 'Era xfj? SoOstaiqc; eu-Mac; 7tETtepaa^Evr]<; xpiycovov laoTtXeupov auaxi^aaCTdai. "Eaxw f] Bo'deToa su'dsla Tt£7tepaa[ievr) r) AB. Aei 8r) era xrjg AB su-Mac; xpiywvov [aoTtXsupov auaxrjaacrdoa. Ksvxpcp jiev xG A 8iaoxr]|jiaxi 8e xG AB xuxXog yEypdcpi&M 6 BrA, xal TtdXiv xsvxpcp [Lev xG B Siaaxr^axi 8e xG BA xuxXoc; yeypa.^-doi 6 ArE, xal diro xou T orjjieiou, xai}' o xsjivouaiv dXXr|Xou<; ol xuxXoi, em xd A, B ar](jieTa £Tie^£U)cd«oav eui&eTai di FA, TB. Kdi etc el xo A or)(ieTov xevxpov eaxl xou TAB xuxXou, lay] eaxlv r) AT xfj AB- TtaXiv, inzi xo B cruie'iov xevxpov eaxl xou IAE xuxXou, Xat] eaxlv r] Br xfj BA. eBeix'dr] 8e xal f) TA xfj AB iar)' exaxepa dpa xGv IA, TB xfj AB eaxiv Tar), xa 8e xG auxG I'oa xal dXXf|Xoic; eaxlv I'aa- xal r) TA dpa xfj TB eaxiv lay]- al xpeu; dpa al TA, AB, Br laai dXXf|Xaic; siaiv. 'IaonXsupov dpa eaxl xo ABr xpiycovov. xal auveaxaxai era xfjc 8oi9e[ar]c; eMeiac; TtETtepaa^iev/jc; xrjg AB. omep e8ei noifjaai. t The assumption that the circles do indeed cut one another should be that two straight-lines cannot share a common segment. Proposition 1 To construct an equilateral triangle on a given finite straight-line. Let AB be the given finite straight-line. So it is required to construct an equilateral triangle on the straight-line AB. Let the circle BCD with center A and radius AB have been drawn [Post. 3], and again let the circle ACE with center B and radius BA have been drawn [Post. 3]. And let the straight-lines CA and CB have been joined from the point C, where the circles cut one another,* to the points A and B (respectively) [Post. 1]. And since the point A is the center of the circle CDB, AC is equal to AB [Def. 1.15]. Again, since the point B is the center of the circle CAE, BC is equal to BA [Def. 1.15]. But CA was also shown (to be) equal to AB. Thus, CA and CB are each equal to AB. But things equal to the same thing are also equal to one another [C.N. 1]. Thus, CA is also equal to CB. Thus, the three (straight- lines) CA, AB, and BC are equal to one another. Thus, the triangle ABC is equilateral, and has been constructed on the given finite straight-line AB. (Which is) the very thing it was required to do. aunted as an additional postulate. There is also an implicit assumption P'- IIp6<; xG 8oil>£vxi or)\izi<x> xfj So-Oslor) eMteiqc Tarjv eu-delav deo-dai. 'Eaxco xo \iev SotJev ar)|isTov xo A, f) Be 8oif)eTaa eu'dela f] Br- 8eT 8fj Ttp6<; xG A arpeta xfj Bo'ddar] eO'deia xfj Br iar]v euiMav deo-dai. 'Erav£eu)c&co yap duo xou A ar)(i£iou era xo B ar)[ieTov su-dela f] AB, xal auveaxdxw etc' auxrjc; xpiyovov ia6n:Xeupov xo AAB, xal expepXf|aT!}«aav etc' cu-deiac; xau; AA, AB Proposition 2" 1 " To place a straight-line equal to a given straight-line at a given point (as an extremity) . Let A be the given point, and BC the given straight- line. So it is required to place a straight-line at point A equal to the given straight-line BC. For let the straight-line AB have been joined from point A to point B [Post. 1], and let the equilateral trian- gle DAB have been been constructed upon it [Prop. 1.1]. 8 ETOIXEIfiN a'. ELEMENTS BOOK 1 eMelai oci AE, BZ, xal xevxpw |iev iu B Biaaxr^iaxi 8e iu Br xuxXoc; yeypdcp-dw 6 THO, xal iraXiv xevxpw xfi A xal 8iaaxr|uaxi iu AH xuxXoc; yeypdcp-dw 6 HKA. E 'Etc! ouv to B ar^elov xevxpov eaxl xoD rHO, lar\ eaxiv f] Br xfj BH. TtdXiv, ETtel xo A arpelov xevxpov eaxl xoO HKA xuxXou, iar) eaxiv f] A A xfj AH, Sv f) A A xfj AB Iar] eaxiv. Xomf] dpa f) AA XoiTtfj xfj BH eaxiv 'iar). e8e[)cdr] 8s xod f] BE xfj BH iar) - exaxepa dpa xfiv AA, Br xfj BH eaxiv iar). xd 8s xw auxw laa xal dXXf|Xoi<; eaxiv laa- xal f) AA dpa xfj Br eaxiv iar). npo<; dpa tw 8oi[)evxi arpeio xo A xfj Bo'deiar] cu-Ma xfj Br iar] euiJeTa xelxai f] AA- omp eSei Ttoirjaai. t This proposition admits of a number of different cases, depending on Euclid invariably only considers one particular case — usually, the most c And let the straight-lines AE and BF have been pro- duced in a straight-line with DA and DB (respectively) [Post. 2] . And let the circle CGH with center B and ra- dius BC have been drawn [Post. 3], and again let the cir- cle GKL with center D and radius DG have been drawn [Post. 3]. F E Therefore, since the point B is the center of (the cir- cle) CGH, BC is equal to BG [Def. 1.15]. Again, since the point D is the center of the circle GKL, DL is equal to DG [Def. 1.15]. And within these, DA is equal to DB. Thus, the remainder AL is equal to the remainder BG [C.N. 3]. But BC was also shown (to be) equal to BG. Thus, AL and BC are each equal to BG. But things equal to the same thing are also equal to one another [C.N. 1]. Thus, AL is also equal to BC. Thus, the straight-line AL, equal to the given straight- line BC, has been placed at the given point A. (Which is) the very thing it was required to do. ie relative positions of the point A and the line BC. In such situations, ficult — and leaves the remaining cases as exercises for the reader. y'- Auo SoOeiafiv eu-deiaiv dviawv duo xrjc; ^ie[£ovo<; xfj eXdaaovi l'ar)v eurMav dcpeXelv. 'Eaxwaav ai Bo-deTaai 8uo euiJeTai dviaoi ai AB, T, 5v ^ei^wv eaxM f) AB- Be! Bf) aTto xfjc; [ieiZ,o\oc, xfjc; AB xfj eXdaaovi xfj T I'arjv eO'delav dcpeXelv. KeiadM npbc, x£> A ar)[ie(w xfj T eu'dela iar] f) AA- xal xevxpo y.ev xo A 5iaaxf|[jiaxi Be xw AA xuxXoc; yeypdcp-dw 6 AEZ. Kod enel to A ar^elov xevxpov eaxl xoO AEZ xuxXou, Proposition 3 For two given unequal straight-lines, to cut off from the greater a straight-line equal to the lesser. Let AB and C be the two given unequal straight-lines, of which let the greater be AB. So it is required to cut off a straight-line equal to the lesser C from the greater AB. Let the line AD, equal to the straight-line C, have been placed at point A [Prop. 1.2]. And let the circle DEF have been drawn with center A and radius AD [Post. 3]. 9 ETOIXEIfiN a'. ELEMENTS BOOK 1 Tar) eaxlv f] AE xfj AA- dXXd xal i\ T xfj A A eaxiv for), exaxepa apa xov AE, T xfj AA eaxiv Tor)- waxe xai f) AE xfj T eaxiv iar). r Auo apa Bcdeiaaiv euiSeifiv dv[o«v xaiv AB, T duo xfjc; (ie[£ovo<; xfj? AB xfj eXdaaovi xfj T iar) dcpfiprjxai f] AE- ojtep eSei noifjaai. 5'. Edv 860 xpiywva xa<; 860 nXeupdc; [xalc] Suol rcXeupau; Taac; e/r] exaxepav exaxepa xal xf]v y^viav xfj ywvia larjv e/rj xf]v Otto xGv ia«v eMeifiv Ttepiexojievrjv, xai T ^ v pdaiv xf) pdaei I'arjv e^ei, xal xo xptywvov xw xpiyovo laov eaxai, xal ai XoiTtal yoviai xdu; XoiTtau; ywviau; laai eaovxai exaxepa exaxepa, Ocp' ac ai laai TtXeupal Gitoxeivouaiv. A A TiaxM 860 xpiytova xd ABr, AEZ xdc; 860 icXeupac; xdc AB, Ar xal<; 8ual TtXeupalg xdu; AE, AZ i'aa<; exovxa exaxepav exaxepa xf]v ^.ev AB xfj AE xf)v 8e Ar xfj AZ xal y«v(av xf]v utco BAT ywvia xfj Otco EAZ larjv. Xeyco, 6x1 xal [3daic; f) Br pdaei xrj EZ Tar) eaxiv, xal xo ABr xpiyovov xw AEZ xpiywvw I'aov eaxai, xal ai Xoiical ytoviai xau; XoiTtalc; ywviau; laai eaovxai exaxepa exaxepa, Ocp'' ac; ai laai TtXeupal UTtoxeivouaiv, f) [ie\ utco ABr xrj utco AEZ, f] 8e utco ArB xrj utco AZE. 'Ecpap|jio^o|jievou yap xou ABr xpiywvou era xo AEZ xpiycovov xal xi'dejievou xou jiev A ar][ieiou era xo A orj^ielov And since point A is the center of circle DEF, AE is equal to AD [Def. 1.15]. But, C is also equal to AD. Thus, and C are each equal to AD. So is also equal to C [C.N. 1]. c B Thus, for two given unequal straight-lines, AB and C, the (straight-line) AE, equal to the lesser C, has been cut off from the greater AB. (Which is) the very thing it was required to do. Proposition 4 If two triangles have two sides equal to two sides, re- spectively, and have the angle (s) enclosed by the equal straight-lines equal, then they will also have the base equal to the base, and the triangle will be equal to the tri- angle, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining an- gles. A D Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF, re- spectively. (That is) AB to DE, and AC to DF. And (let) the angle BAG (be) equal to the angle EDF. I say that the base BC is also equal to the base EF, and triangle ABC will be equal to triangle DEF, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles. (That is) ABC to DEF, and ACB to DFE. For if triangle ABC is applied to triangle DEF,^ the point A being placed on the point D, and the straight-line 10 ETOIXEIfiN a'. ELEMENTS BOOK 1 xrjc 8e AB zu'dziac, era xrjv AE, ecpapjioaei xal to B arpeiov era to E 8ia to Taiqv elvai x/jv AB xrj AE- ecpap|ioadaT]c; 8f) xfjt; AB em TTjv AE ^apjioaei xal f) Ar eMela era xr]v AZ 6id to ToTjv elvai xr)v utio BAr ywviav xfj utio EAZ' Saxe xal to r or)(ieIov era to Z arpelov ecpapjioaei 5ia to larjv TidXiv elvai t/)v Ar xfj AZ. dXXa [ir\\> xal xo B era xo E ecprjp^oxei- Saxe pdoic; f] Br era pdaiv xrjv EZ ecpapjioaei. el yap xoO (iev B era xo E ecpapjioaavxog xoO Be T era xo Z f) Br pdau; era xr)v EZ oux ecpapjioaei, 860 eui&eTai /wpiov uepie^ouoiv ojiep eaxlv dBuvaxov. ecpapjioaei dpa f] Br pdai<; era x/]v EZ xal Tar) auxfj eaxai- Saxe xal oXov xo ABr xpiytovov era oXov xo AEZ xpiywvov ecpapjioaei xal I'aov auxcp eaxai, xal ai Xomal ycoviai era xdg Xoina? ycoviag ecpapjioaouai xal laai auxalc eaovxai, f) ^ev utio ABr xrj utio AEZ fj Be utio ArB xrj utio AZE. 'Eav dpa 860 xpiyiova xac; 8uo TiXeupag [xdig] 660 TiXeupdic; iaac; evt) exaxepav exaxepa xal xrjv yioviav xrj ywvia lorjv exn xrjv utio xfiv Tacov eui5eifiv Tiepie/o|jievr]v, xal xr]v pdaiv xf) pdaei Tarjv e<;ei, xal xo xpiywvov xG xpiycovco I'aov eaxai, xal ai Xouial ycovlai xalg Xoiraiu; ycoviaig laai eaovxai exaxepa exaxepa, ucp' a<; al laai TiXeupal UTioxeivouaiv oTiep eBei BeT^ai. AB on DE, then the point B will also coincide with E, on account of AB being equal to DE. So (because of) ^4i? coinciding with DE, the straight-line AC will also coincide with DF, on account of the angle BAG' being equal to EDF. So the point C will also coincide with the point F, again on account of AC being equal to DF. But, point B certainly also coincided with point E, so that the base BC will coincide with the base EF. For if B coin- cides with E, and C with F, and the base -BC does not coincide with EF, then two straight-lines will encompass an area. The very thing is impossible [Post. 1].* Thus, the base BC will coincide with EF, and will be equal to it [C.N. 4]. So the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it [C.N. 4]. And the remaining angles will coincide with the remain- ing angles, and will be equal to them [C.N. 4]. (That is) ABC to DEF, and ACB to DFE [C.N. 4]. Thus, if two triangles have two sides equal to two sides, respectively, and have the angle (s) enclosed by the equal straight-line equal, then they will also have the base equal to the base, and the triangle will be equal to the tri- angle, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining an- gles. (Which is) the very thing it was required to show. t The application of one figure to another should be counted as an additional postulate, t Since Post. 1 implicitly assumes that the straight-line joining two given points is unique. S . Tc5v iaoaxeXGv xpiywvwv ai xpog xfj pdaei ywviai iaai dXXrjXau; eiaiv, xal TipoaexpXr]i9eia5v xwv lacov eui9ei«v ai utio x/]v pdaiv ywviai laai aXXrjXait; eaovxai. A Tiaxco xpiywvov iaoaxeXec; xo ABr iar]v 'iyov xr]v AB TiXeupav xfj AT TiXeupa, xal TipoaexpepXfja'dwaav en eCWteiac; xdi<; AB, Ar einMai ai BA, TE- Xeyw, oxi f] [ie\> utio ABr ywvia xfj utio ArB Tar) eaxiv, f] Be Otto TBA xrj utio BrE. EiXrjcp'dco yap era xfjc BA xu/6v ar^ieiov xo Z, xal dcprjpria'dco arco xfj? [iei£ovo<; xfjc; AE xfj eXdaaovi xfj AZ Proposition 5 For isosceles triangles, the angles at the base are equal to one another, and if the equal sides are produced then the angles under the base will be equal to one another. A D E Let ABC be an isosceles triangle having the side AB equal to the side AC, and let the straight-lines BD and CE have been produced in a straight-line with AB and AC (respectively) [Post. 2]. I say that the angle ABC is equal to ACB, and (angle) CBD to BCE. For let the point F have been taken at random on BD, and let AG have been cut off from the greater AE, equal 11 ETOIXEIfiN a'. ELEMENTS BOOK 1 Tar] f] AH, xai £Tce£eu)(iL>Gjaav ai Zr, HB sMeiai. 'EtceI ouv iar] eaxlv f] [iev AZ xfj AH f] 8e AB xfj Ar, 8uo 8f| al ZA, Ar Sua! xdic. HA, AB laai elolv exaxepa sxaxspor xal ywviav xoivrjv Tcepii)(ouai xf)v utco ZAH' pdaic, apa f) Zr pdaei xfj HB Tar) eaxiv, xal xo AZr xpiyovov iu AHB xpiyovo Taov eaxai, xal ai Xomal ycoviai toac. XomaTc. ycoviaic, I'oai eaovxai exaxepa exaxepa, ucp' ac, ai Taai TcXeupal UTtoxeivouaiv, r] ^ev utco ATZ xfj utco ABH, f) 8e utco AZr xfj utco AHB. xal CTcel 6Xt] r] AZ oXy] xfj AH eaxiv Tar], Gv f) AB xfj Ar eaxiv Tar], XoiTcf] apa f] BZ Xomfj xfj TH eaxiv larj. eBeix^r] Se xal f] Zr xfj HB for) - Suo 8f] ai BZ, Zr 8ual xdic. TH, HB Taai eiaiv exaxepa exaxepa- xal yiovia f] utco BZr yiovia xrj utco THB lor), xal pdaic auxcov xoivf] f) BT- xal xo BZr apa xpiywvov x« THB xpiycbvcp Taov eaxai, xal ai XoiTtal ywviai xdic XoiTcdic ycoviaic Taai eaovxai exaxepa exaxepa, ucp' ac ai i'oai TcXeupal UTioxeivouaiv I'or) dpa eaxlv f) ^iev utco ZBr xfj utto HTB f] Se utco BrZ xfj Gtco TBH. etxeI ouv 6Xt) f] utco ABH ytovia oXr) xfj utco ArZ ytovia £8ei)cdr) iar), cov f] utco TBH xfj utco BrZ lar], Xoiti/) apa f] utco ABr XoiTxfj xfj utco ArB screw iar) - xai eiai Tcpoc xfj pdaei xou ABr xpiytovou. eSei/'dr] 8s xal f] utco ZBr xfj utco HrB Tar)- xai eiaiv utco xf)v pdaiv. Tcov dpa iaoaxeXGv xpiytovtov ai xpoc xfj pdaei ycoviai Taai dXXfjXaic eiaiv, xai TcpoaexpXrji&eiaGv xfiv Tacov euifteicov ai utco xf]v pdaiv ycoviai Taai dXXrjXaic eaovxai- OTcep eSei 8eT^ai. 9 . 'Edv xpiycovou ai 8uo ycoviai Taai dXXf]Xaic coaiv, xal ai utco xdc Taac ycoviac UTCoxeivouaai TcXeupal Taai dXXr]Xaic eaovxai. A 'Eaxco xpiycovov xo ABr Tarjv e)(ov xrjv utco ABr ycoviav xfj utco ArB ycovia- Xeyo, oxi xal TcXeupd f] AB TcXeupa xfj Ar eaxiv iar). to the lesser AF [Prop. 1.3]. Also, let the straight-lines FC and GB have been joined [Post. 1]. In fact, since AF is equal to AG, and AB to AC, the two (straight-lines) FA, AC are equal to the two (straight-lines) GA, AB, respectively. They also encom- pass a common angle, FAG. Thus, the base FC is equal to the base GB, and the triangle AFC will be equal to the triangle AGB, and the remaining angles subtendend by the equal sides will be equal to the corresponding remain- ing angles [Prop. 1.4]. (That is) ACF to ABG, and AFC to AGB. And since the whole of AF is equal to the whole of AG, within which AB is equal to AC, the remainder BF is thus equal to the remainder CG [C.N. 3]. But FC was also shown (to be) equal to GB. So the two (straight- lines) BF, FC are equal to the two (straight-lines) CG, GB, respectively, and the angle BFC (is) equal to the angle CGB, and the base BC is common to them. Thus, the triangle BFC will be equal to the triangle CGB, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles [Prop. 1.4]. Thus, FBC is equal to CGB, and BCF to CBG. There- fore, since the whole angle ABG was shown (to be) equal to the whole angle ACF, within which CBG is equal to BCF, the remainder ABC is thus equal to the remainder ACB [C.N. 3]. And they are at the base of triangle ABC. And FBC was also shown (to be) equal to GCB. And they are under the base. Thus, for isosceles triangles, the angles at the base are equal to one another, and if the equal sides are produced then the angles under the base will be equal to one an- other. (Which is) the very thing it was required to show. Proposition 6 If a triangle has two angles equal to one another then the sides subtending the equal angles will also be equal to one another. A Let ABC be a triangle having the angle ABC equal to the angle ACB. I say that side AB is also equal to side AC. 12 ETOIXEIfiN a'. ELEMENTS BOOK 1 El yap aviao<; eaxiv f] AB xfj Ar, rj exepa aOxCSv ^.el^tov eaxlv. eaxo (jlel^cov f\ AB, xal dcprjpfia'do dra 5 ) xfj? ^el^ovoc; xfj? AB xfj eXdxxovi xfj Ar lot) f) AB, xal eTie£eu)(i&Gj ^ AT. 'End ouv Xat] eaxlv f] AB xrj Ar xoivf] 8s r) Br, 8uo 8f] at AB, Br 8uo xau; Ar, TB I'aai elalv exaxepa exaxepa, xal ycovla f] Otto ABr ycovla xfj utio ArB eaxiv I'ar)- pdaic; dpa #j Ar pdaei xrj AB lor] eaxlv, xal xo ABr xplycovov xfi ArB xpiycovco laov eaxai, xo eXaaaov xfi [isl^ovi - oTiep dxoTtov oux dpa dviaoc; eaxiv f) AB xfj Ar - I'ar) dpa. 'Edv dpa xpiycovou al 6uo ycovlai laai dXXrjXau; Saiv, xal at Otio xdc; laac; ycovlac; UTtoxelvouaai TtXeupal laai dXXrjXau; eaovxau OTiep eSei Set^ai. For if AB is unequal to AC then one of them is greater. Let AB be greater. And let DB, equal to the lesser AC, have been cut off from the greater AB [Prop. 1.3]. And let DC have been joined [Post. 1]. Therefore, since DB is equal to AC, and BC (is) com- mon, the two sides DB, BC are equal to the two sides AC, CB, respectively, and the angle DBC is equal to the angle ACB. Thus, the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB [Prop. 1.4], the lesser to the greater. The very notion (is) absurd [C.N. 5]. Thus, AB is not unequal to AC. Thus, (it is) equals Thus, if a triangle has two angles equal to one another then the sides subtending the equal angles will also be equal to one another. (Which is) the very thing it was required to show. t Here, use is made of the previously unmentioned common notion that if two quantities are not unequal then they must be equal. Later on, use is made of the closely related common notion that if two quantities are not greater than or less than one another, respectively, then they must be equal to one another. 'Era xrjc; auxfjc eui!)ela<; 8uo xau; auxale; eO'delau; aXXai 8uo euiJkTaL iaai exaxepa exaxepa ou auaxai3f]aovxai Ttpoc; aXXcp xal aXXco arpeltp era xd auxd fiepr] xd auxd Tiepaxa exouaai xau; z\ dp/fjc; euiitelau;. El yap Suvaxov, era iff auxfji; eMelac; iff AB 8uo xdic; auxau; eCcdelau; xau; AT, TB dXXai 8uo eu'deTai at AA, AB laai exaxepa exaxepa auveaxdxoaav Tipoc; aXXcp xal dXXcp ay][ie'icd i& is r xal A era xd auxd ^eprj xd auxd Tiepaxa e/ouaai, coaxe lar\v eTvai xf)v [Lev IA xfj AA xo auxo rcepac; e/ouaav auxfj xo A, xf)v 8e TB xrj AB xo auxo Tiepac; exou- aav auxfj xo B, xal eTieCeuy^dw ^ TA. 'Erad ouv Xar\ eaxlv r\ AT xfj AA, Tar) tail xal ycovla f] utio ATA xfj utio AAT- ^tel^cov dpa f] utio AAT xrjc; utio ATB- tioXXco dpa f] utio TAB ueli^cov eaxl xfj? utio ATB. TcdXiv STtel Tar] eaxlv f) TB xfj AB, Tar] eaxl xal ycovla f) utio TAB ycovla xfj utio ATB. eSelx'dr) 8e auxrjc xal TtoXXcp HelC^cov oTiep eaxlv d8uvaxov. Oux dpa era xrjc auxfjc eui9elac. 86o xau; auxdu; eu'delau; Proposition 7 On the same straight-line, two other straight-lines equal, respectively, to two (given) straight-lines (which meet) cannot be constructed (meeting) at a different point on the same side (of the straight-line), but having the same ends as the given straight-lines. For, if possible, let the two straight-lines AC, CB, equal to two other straight-lines AD, DB, respectively, have been constructed on the same straight-line AB, meeting at different points, C and D, on the same side (of AB), and having the same ends (on AB). So CA is equal to DA, having the same end A as it, and CB is equal to DB, having the same end B as it. And let CD have been joined [Post. 1]. Therefore, since AC is equal to AD, the angle ACD is also equal to angle ADC [Prop. 1.5]. Thus, ADC (is) greater than DCB [C.N. 5]. Thus, CDB is much greater than DCB [C.N. 5]. Again, since CB is equal to DB, the angle CDB is also equal to angle DCB [Prop. 1.5]. But it was shown that the former (angle) is also much greater 13 ETOIXEIfiN a'. ELEMENTS BOOK 1 dXXai 8uo eui5eTauaai exaxepa exaxepa auaTocdfjaovToa Ttpo? dXXcp xal dXXw arj^eio era xd aOxd [Jiepr] xd auxd Ttepaxa e/ouaai xal? 15 dp/f)? eOiiteiai?- oitep e8ei BeT^ai. 'Edv 8uo xpiycova xd? 860 iiXeupd? [xal?] 860 TiXeupal? laa? l)(T] exaxepav exaxepa, zyr\ 8e xal xrjv pdaiv xfj pdaei larjv, xal xrjv ywviav xfj yiovia I'arjv eijei xf]v unb xwv Tacov eMeiwv Ttepiexojiev/jv. 15axw 860 xpiytova xd ABr, AEZ xd? Buo TtXeupd? xd? AB, Ar xai? 660 nXeupaT? xal? AE, AZ !aa? exovxa exaxepav exaxepa, xrjv [lev AB xfj AE xf]v 8s Ar xfj AZ- 8e xal pdaiv xrjv Br pdaei xfj EZ l'ar]v Xey«, oxi xal ycovia f) Ono BAr ywvia xfj 6716 EAZ eaxiv iar). 'E(fctp\ioZo[Levo\j yap xou ABr xpiywvou era xo AEZ xpiyovov xal xide^tevou xou \±ev B af][ie'iou era xo E arpeTov xfj? Be Br euT!)e[a? era x/]v EZ ecpapjioaei xal xo T arpelov excl xo Z Bid xo iar]v elvai xf]v Br xfj EZ- ecpapjioadar]? 8rj xfj? Br Ira xrjv EZ ecpap^toaouai xal al BA, TA era xd? EA, AZ. ei yap pdai? [lev f\ BT era pdaiv xf)v EZ ecpap^toaei, ai 8e BA, Ar TiXeupal era xd? EA, AZ oux ecpapjioaouaiv dXXd TtapaXXd^ouaiv cb? ai EH, HZ, auaxa'driaovxai era xfj? auxfj? euiJeia? 860 xaT? auxaT? eu-deiai? dXXai 860 eu-delai laai exaxepa exaxepa rcpo? aXXo xal dXXto ar^eiw era xd auxa [ispf] xd auxa Ttepaxa s/ouaai. ou auviaxavxai Be- oux dpa ecpap^oi^o^evr)? xfj? Br pdaew? era xrjv EZ pdaiv oux ecpapuoaouai xal ai BA, Ar TiXeupal era xd? EA, AZ. ecpapjioaouaiv dpa- waxe xal ywvia f] Otco BAr era ywviav xf|V Oko EAZ ecpap^toaei xal iar) auxfj eaxai. 'Edv dpa Buo xpiywva xd? Suo iiXeupd? [xal?] 860 TiXeupal? I'aa? e^T] exaxepav exaxepa xal xrjv pdaiv xfj pdaei lar]v e^T), xal xf]v ytoviav xfj yovia ia/]v e^ei xf)v utco xwv lawv eu-deiwv Ttepiexo^evrjv oTtep eBei SeT^ai. (than the latter). The very thing is impossible. Thus, on the same straight-line, two other straight- lines equal, respectively, to two (given) straight-lines (which meet) cannot be constructed (meeting) at a dif- ferent point on the same side (of the straight-line), but having the same ends as the given straight-lines. (Which is) the very thing it was required to show. Proposition 8 If two triangles have two sides equal to two sides, re- spectively, and also have the base equal to the base, then they will also have equal the angles encompassed by the equal straight-lines. Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF, respectively. (That is) AB to DE, and AC to DF. Let them also have the base BC equal to the base EF. I say that the angle BAC is also equal to the angle EDF. For if triangle ABC is applied to triangle DEF, the point B being placed on point E, and the straight-line BC on EF, then point C will also coincide with F, on account of BC being equal to EF. So (because of) BC coinciding with EF, (the sides) BA and CA will also co- incide with ED and DF (respectively) . For if base BC coincides with base EF, but the sides AB and AC do not coincide with ED and DF (respectively), but miss like EG and GF (in the above figure), then we will have con- structed upon the same straight-line, two other straight- lines equal, respectively, to two (given) straight-lines, and (meeting) at a different point on the same side (of the straight-line), but having the same ends. But (such straight-lines) cannot be constructed [Prop. 1.7]. Thus, the base BC being applied to the base EF, the sides BA and AC cannot not coincide with ED and DF (respec- tively). Thus, they will coincide. So the angle BAC will also coincide with angle EDF, and will be equal to it [C.N. 4]. Thus, if two triangles have two sides equal to two side, respectively, and have the base equal to the base, 14 ETOIXEIfiN a'. ELEMENTS BOOK 1 then they will also have equal the angles encompassed by the equal straight-lines. (Which is) the very thing it was required to show. 0'. Tf|v Softeiaav yioviav eMuypa^ov 5(xa xe^elv. A 'Eaxco f] So'deiaa y«v(a eO'duypa^oc; f] hub BAr. 8eT 8r) auxfjv Si^a xejieTv. EtXfiqydw iiti xfjc AB xu^ov GTj^ielov to A, xal dcpr)pf|C7d« dno Tfjc; Ar xfj AA Tar) rj AE, xal eite^eux^M f) AE, xal auveaxaxo) em xfjc AE xpiywvov taoTtXeupov xo AEZ, xal eTisCeux'dw f] AZ- Xeyto, 6x1 f\ O716 BAr ycovia Bl^a xexurjxai O716 xfjc AZ eu'ddac. Tkel yap for] eaxlv f] AA xfj AE, xoivf) 8e f\ AZ, 860 8f) at AA, AZ Sual xau; EA, AZ foai eiatv exaxepa exaxepa. xal pdai<; f] AZ pdaei xfj EZ far) eaxtv ywvia apa r\ bub AAZ ywvia xfj O716 EAZ for] eaxtv. TL apa 5odefoa yovta eiWJuypa^ot; fj O716 BAr 8txa xex^.r)xai utco xfjc AZ eu'detac;- onep e8ei uoifjaai. Proposition 9 To cut a given rectilinear angle in half. A Let BAC be the given rectilinear angle. So it is re- quired to cut it in half. Let the point D have been taken at random on AB, and let AE, equal to AD, have been cut off from AC [Prop. 1.3], and let DE have been joined. And let the equilateral triangle DEF have been constructed upon DE [Prop. 1.1], and let AF have been joined. I say that the angle BAC has been cut in half by the straight-line AF. For since AD is equal to AE, and AF is common, the two (straight-lines) DA, AF are equal to the two (straight-lines) EA, AF, respectively. And the base DF is equal to the base EF. Thus, angle DAF is equal to angle EAF [Prop. 1.8]. Thus, the given rectilinear angle BAC has been cut in half by the straight-line AF. (Which is) the very thing it was required to do. l . Tf]v Sodefoav eu^ETav Tcenepaajievrjv 8i);a xejieTv. Tiaxa> f] Sodefoa eO'dela Keiiepaa^evr) f) AB- 8eT 8f] xf]v AB euiMav TteTtepaauevrjv Btya xe^ielv. Suveaxdxo en auxfjc; xptywvov taoTtXeupov xo ABT, xal xex^ir|<yda> f) Gtco ATB yovta 8i)(a xfj EA euifteta- Xeyw, oxi f] AB eu'dela 8txa xex^rjxai xaxd xo A ar][ieTov. Tkel yap Tar) eaxlv f\ AT xfj TB, xoivf] 8s f] TA, 860 8f] at AT, TA 860 xau; BT, TA foai eiatv exaxepa exaxepa- xal ycovta f] O716 ATA ywvta xfj O716 BTA for] eaxtv pdau; apa Proposition 10 To cut a given finite straight-line in half. Let AB be the given finite straight-line. So it is re- quired to cut the finite straight-line AB in half. Let the equilateral triangle ABC have been con- structed upon (AB) [Prop. 1.1], and let the angle ACB have been cut in half by the straight-line CD [Prop. 1.9]. I say that the straight-line AB has been cut in half at point D. For since AC is equal to CB, and CD (is) common, 15 ETOIXEIfiN a'. ELEMENTS BOOK 1 f) AA (3aa£i xfj BA far) eaxiv. 'H apa Scdelaa euifteTa TiETtepaa^evr] f\ AB 8[/a xexurjxai xaxa to A- oracp eSei rcoifjaai. the two (straight-lines) AC, CD are equal to the two (straight-lines) BC, CD, respectively. And the angle ACD is equal to the angle BCD. Thus, the base AD is equal to the base BD [Prop. 1.4]. c Thus, the given finite straight-line AB has been cut in half at (point) D. (Which is) the very thing it was required to do. ia'. Tfj 8o$£[ar) sMeia dno tou npbc, auxrj BoiSevxoc; ar^dou npbq 6pM<; ywviai; eMeiav ypa^rjv dyayeiv. z ATE 'Eaxco f] ^lev BoiSefaa euiiteTa f) AB to 8e bo-&e\ af][ieiov in' auxrjg to T- 8eT Srj arco tou T ar][ieio\j xfj AB cu-Ma npoc opMg ytoviac; sMeTav ypaji|if]v dyayeiv. EiX^-dtd era xrjc; Ar xu)(6v a/jjislov to A, xal xeia-du if) IA Tor) f) TE, xal auvsaxdxto era xrjc AE xpiywvov iooTiXeupov xo ZAE, xal inzt.zux'Qid f] ZE Xeyto, oxi xfj SoiEteiar) EUifrda xfj AB duo tou upoc auxfj 8oi9£vto<; ar)[idou tou r Tipoc; op-dac; yioviac; euiSsIa ypajijif) rjxxai f\ ZT. 'End yap far) eaxlv f) AT xfj TE, xoivf) 8e f] TZ, 8uo 8f) ai AT, TZ 8ual xal<; ET, TZ faai dalv exaxspa sxaxepa- xal pdaig T) AZ pdaeri xfj ZE lor) eaxiv ywvia apa f) uko ATZ ywvta xfj (mo ETZ far] eaxtv xai efaiv £cpe^fj<;. oxav 8e eu'dda Ik' eiWteTav axaiftefaa xa<; ecpec;fj<; ywv[a<; faa<; dXXf)Xaic rcoifj, op'df] exaxspa xwv fawv ywviwv eaxiv op'df) apa eaxlv exaxepa xwv utio ATZ, ZTE. Proposition 11 To draw a straight-line at right-angles to a given straight-line from a given point on it. F D C E Let AB be the given straight-line, and C the given point on it. So it is required to draw a straight-line from the point C at right-angles to the straight-line AB. Let the point D be have been taken at random on AC, and let CE be made equal to CD [Prop. 1.3], and let the equilateral triangle FDE have been constructed on DE [Prop. 1.1], and let EC have been joined. I say that the straight-line EC has been drawn at right-angles to the given straight-line AB from the given point C on it. For since DC is equal to CE, and CF is common, the two (straight-lines) DC, CF are equal to the two (straight-lines), EC, CF, respectively. And the base DF is equal to the base FE. Thus, the angle DCF is equal to the angle ECF [Prop. 1.8], and they are adjacent. But when a straight-line stood on a(nother) straight-line 1G ETOIXEIfiN a'. ELEMENTS BOOK 1 Tfj apa BoiDeior] euiSeia xfj AB and xoO Ttp6<; auxrj BcedevTcx; ar)^e(ou tou T Tcpoc; dpfiuc, ywviac; eO'deTa ypa^r) rjxxai f] TZ- oitep eBei noirjaai. 10'. 'Era x/]v 8oi!)eTaav eu'deuxv aneipov &tco xoO So-devxoc arj^eiou, 8 [ir\ eaxiv etc' auxfj<;, xdiJexov euiMav ypa^ir]v dyayeTv. z "Eaxo f] ^tev So'de'iaa eu'de'ia aiceipcx; f] AB xo 8e 8oi9ev ar)^eTov, 8 ^ eaxiv etc' auxfj<;, xo E 8eT Br] era xrjv Bo-deTaav eMelav dneipov xr]v AB duo xoO Scdevxot; ar)|J.e[ou xoO T, 8 \ir\ eaxiv etc' auxfjc;, xdcdexov eu-delav ypa^rjv dyayeTv. EiXr]cpdco yap tiii xd exepa jiepr] xfj<; AB eu^eiac; xu^ov arjjieTov xo A, xdi xevxpw [lev xai T 8iaaxr]jjiaxi 8e xa> EA xuxXoc; yeypdcpiJw 6 EZH, xdi xex|ir|ada) f] EH eui5ela 8i/a xaxd xo 6, xdi C7te£eu)(iL)Maav di TH, TO, TE euiiteTai- Xeya>, oxi era xr)v BcdeTaav euaDeTav aracipov xrjv AB duo xou Bo-devxoc; at]\ieiou xou T, 3 \Lr\ eaxiv sit' auxrjc;, xd-dexoc; rjxxai f] TO. 'Eracl ydp Xar\ eaxiv f) HO xfj 8E, xoivr] Be f) Or, Buo 8/) di H0, OT Buo xau; EO, Or Taai eialv exaxepa exaxepa- xdi pdaic; f) TH [3daei xfj TE eaxiv lot]- yiovia apa f) Otto T0H ya>v[a xrj Otto E0r eaxiv Tar), xai eiaiv ecpe^fjg. oxav Be eO'deTa en' eui5eTav axa-delaa xdc; ecpe^rjc ywviac; Taag dXXrjXaic; Ttoifj, 6p$r) exaxepa xwv Tacov ycoviwv eaxiv, xai f) ecpeaxrjxula eui5ela xd-dexog xaXelxai ecp' fjv ecpeaxrjxev. 'Era xrjv Bo-deTaav apa eu-delav aracipov xrjv AB duo xou Boi&evxoc; ar\[ieiou xou T, o ^fj eaxiv in auxfjc, xd-dexoc; rjxxai f] TO- oitep eBei icoifjaai. makes the adjacent angles equal to one another, each of the equal angles is a right-angle [Def. 1.10]. Thus, each of the (angles) DCF and FCE is a right-angle. Thus, the straight-line CF has been drawn at right- angles to the given straight-line AB from the given point C on it. (Which is) the very thing it was required to do. Proposition 12 To draw a straight-line perpendicular to a given infi- nite straight-line from a given point which is not on it. F D Let AB be the given infinite straight-line and C the given point, which is not on (AB). So it is required to draw a straight-line perpendicular to the given infinite straight-line AB from the given point C, which is not on {AB). For let point D have been taken at random on the other side (to C) of the straight-line AB, and let the circle EFG have been drawn with center C and radius CD [Post. 3], and let the straight-line EG have been cut in half at (point) H [Prop. 1.10], and let the straight- lines CG, CH, and CE have been joined. I say that the (straight-line) CH has been drawn perpendicular to the given infinite straight-line AB from the given point C, which is not on (AB) . For since GH is equal to HE, and HC (is) common, the two (straight-lines) GH, HC are equal to the two (straight-lines) EH, HC, respectively, and the base CG is equal to the base CE. Thus, the angle CHG is equal to the angle EHC [Prop. 1.8], and they are adjacent. But when a straight-line stood on a(nother) straight-line makes the adjacent angles equal to one another, each of the equal angles is a right-angle, and the former straight- line is called a perpendicular to that upon which it stands [Def. 1.10]. Thus, the (straight-line) CH has been drawn perpen- dicular to the given infinite straight-line AB from the 17 ETOIXEIfiN a'. ELEMENTS BOOK 1 'Eav su'dsla in eMsiav aiadzica. ywviac; Tioifj, fjxoi 8uo dp'&a.z fj 8uatv op-ddic; Taa<; TioirjaEi. a b r EuiJeTa yap xi? f] AB etc' eu-deiav xrjv TA axaiMaa yov(a<; tcoieixco xdc utio TBA, ABA- Xsy", oxi at utio TBA, ABA ycovtai rjxoi 860 op'dai Eiaiv fj Sualv opiDau; I'aai. EE [ie\ ouvTar) saxlv f] utio TBA xfj utio ABA, Suo 6p$a[ Eiaiv. si 8s ou, f\x$(i> aTio xou B arj^isiou xfj TA [sMsia] npbc, opiDat; f) BE' at apa utio TBE, EBA 860 op'dai Eiaiv- xal etieI f] utio TBE Sua! xdic utio TBA, ABE I'ar) saxlv, xoivr] Ttpoaxsia'dw f) utio EBA- at apa utio TBE, EBA xpial xau; utio TBA, ABE, EBA I'aai Eiaiv. TidXiv, etcsi f) utio ABA Sua! xau; utio ABE, EBA Tar) saxlv, xoivr) Tipoaxsla'dw f) utio ABE at apa utio ABA, ABr xpial xau; utio ABE, EBA, ABr Taai siaiv. sSslxi&rjaav 8s xal at utio TBE, EBA xpial xdic; auxaii; Taai- xa 8s xo auxo Taa xal dXXr]Xou; saxlv Taa- xal at utio TBE, EBA dpa xau; utio ABA, ABr Taai siaiv- dXXd at utio TBE, EBA Suo op'dai eiaiv xal at utio ABA, ABr dpa Sualv opiJau; Taai Eiaiv. 'Eav apa su-dsla in suiiteTav axaiSsTaa ywvlac; Tioifj, f)xoi Suo op'ddi; fj 6ualv opiSdic; Taac; TioirjaEi- oTisp sSsi 5sTc;ai. given point C, which is not on ( AB) . (Which is) the very thing it was required to do. Proposition 13 If a straight-line stood on a(nother) straight-line makes angles, it will certainly either make two right- angles, or (angles whose sum is) equal to two right- angles. D B C For let some straight-line AB stood on the straight- line CD make the angles CBA and ABD. I say that the angles CBA and ABD are certainly either two right- angles, or (have a sum) equal to two right-angles. In fact, if CBA is equal to ABD then they are two right-angles [Def. 1.10]. But, if not, let BE have been drawn from the point B at right-angles to [the straight- line] CD [Prop. 1.11]. Thus, CBE and EBD are two right-angles. And since CBE is equal to the two (an- gles) CBA and ABE, let EBD have been added to both. Thus, the (sum of the angles) CBE and EBD is equal to the (sum of the) three (angles) CBA, ABE, and EBD [C.N. 2]. Again, since DBA is equal to the two (an- gles) DBE and EBA, let ABC have been added to both. Thus, the (sum of the angles) DBA and ABC is equal to the (sum of the) three (angles) DBE, EBA, and ABC [C.N. 2]. But (the sum of) CBE and EBD was also shown (to be) equal to the (sum of the) same three (an- gles) . And things equal to the same thing are also equal to one another [C.N. 1]. Therefore, (the sum of) CBE and EBD is also equal to (the sum of) DBA and ABC. But, (the sum of) CBE and EBD is two right-angles. Thus, (the sum of) ABD and ABC is also equal to two right-angles. Thus, if a straight-line stood on a(nother) straight- line makes angles, it will certainly either make two right- angles, or (angles whose sum is) equal to two right- angles. (Which is) the very thing it was required to show. 18 ETOIXEIfiN a'. ELEMENTS BOOK 1 18'. 'Eav Tcpoc; tlvi sMsia xal xG Tcpoc auxfj arista 860 su-dslai [L7] era xa auxa [iepf] xsi^isvai xag scpsi;fjc; ycovia? 8uolv opiSalg iaa<; rcoiScnv, en' sui)siac; saovxai dXXfjXau; ai su-dslai. A E r b A LTpoc; yap xivi su-dsia xfj AB xai xco upog auxfj arjjjieLtp x£> B 860 su-dslai ai Br, BA [if] era xa auxa (icpr) xeijievai xa<; ecpe^fj<; y«via<; xa<; utco ABr, ABA 860 op-dau; laac; Tioieixwaav Xsy«, oxi etc' euiSetai; eaxi xfj TB f] BA. El yap sqxi xfj Br etc' su-Ssiac; f) BA, eax« xfj TB etc' su-dsia^ f] BE. 'Etcei ouv su-dsla f) AB etc' su-dslav xf]v TBE scp£axr]xsv, ai apa utco ABr, ABE ytoviai Suo op-dau; laai siaiv siai 8s xai ai utco ABr, ABA 860 opiSaTi; foai - ai apa utco TBA, ABE xaT? utco TBA, ABA laai eiaiv. xoivf] dcpr)pf|a , dw f) utco TBA- Xoitct) apa f] utco ABE XoiTcfj xfj utco ABA saxiv lay], i] sXaaawv xfj [is[£ovr oTcsp saxiv dSuvaxov. oux dpa etc' sMsiac; saxiv f] BE xfj TB. ojioiok 8f] 8sic;o^£v, oxl ouSs dXXr) xu; TcXfjv xfjc; BA- etc' su-dsiac; dpa saxiv f) TB xfj BA. 'Eav apa Tcpog xivi sui5sia xai cfi Tcpoc; auxfj arista 860 su^siai [iv] sra auxa [tept] xsipisvai xdg scps^fjg yioviac; Suoiv op-ddig Taac; Tcouoaiv, etc' su-dsiac saovxai dXXfjXaic; ai su-dslai- oTcsp sSsi Ssl^ai. is'. 'Eav 860 sO'ds'iai xsjivwaiv dXXf|Xac;, xac; xaxa xopucpfjv ycovia<; laag dXXfjXau; tcoiouoiv. Proposition 14 If two straight-lines, not lying on the same side, make adjacent angles (whose sum is) equal to two right-angles with some straight-line, at a point on it, then the two straight-lines will be straight-on (with respect) to one an- other. A E C B D For let two straight-lines BC and BD, not lying on the same side, make adjacent angles ABC and ABD (whose sum is) equal to two right-angles with some straight-line AB, at the point B on it. I say that BD is straight-on with respect to CB. For if BD is not straight-on to BC then let BE be straight-on to CB. Therefore, since the straight-line AB stands on the straight-line CBE, the (sum of the) angles ABC and ABE is thus equal to two right-angles [Prop. 1.13]. But (the sum of) ABC and ABD is also equal to two right- angles. Thus, (the sum of angles) CBA and ABE is equal to (the sum of angles) CBA and ABD [C.N. 1]. Let (an- gle) CBA have been subtracted from both. Thus, the re- mainder ABE is equal to the remainder ABD [C.N. 3], the lesser to the greater. The very thing is impossible. Thus, BE is not straight-on with respect to CB. Simi- larly, we can show that neither (is) any other (straight- line) than BD. Thus, CB is straight-on with respect to BD. Thus, if two straight-lines, not lying on the same side, make adjacent angles (whose sum is) equal to two right- angles with some straight-line, at a point on it, then the two straight-lines will be straight-on (with respect) to one another. (Which is) the very thing it was required to show. Proposition 15 If two straight-lines cut one another then they make the vertically opposite angles equal to one another. 19 ETOIXEIftN a'. ELEMENTS BOOK 1 Auo yap eu"delai at AB, TA xejivexwaav dXXf|Xac; xaxa to E arjueiov Xeyco, 8x1 iar) eaxlv rj (iev utio AEr ycovta xfj utto AEB, f) 8e utto TEB xfj Otto AEA. A B Tkel yap eO-dsTa f\ AE eu^slav xfjv TA ecpeaxrjxe y«v(a<; Tioiouaa xdc; Otto TEA, AEA, at apa utio TEA, AEA yovtai 8ualv dp'&aa.q I'aai eiaiv. TtdXiv, ercel eui5ela f] AE eit' eMelav xf]v AB ecpeaxrjxe yioviag noiouaa xdg utio AEA, AEB, at apa utio AEA, AEB ycoviai 8ualv opiate; Xaai etatv. e8d)edr)o-av 8e xal at Otto TEA, AEA Buotv opiate; Taai- at apa utio TEA, AEA xau; utio AEA, AEB laai slow, xoivrj d(pr)pf|adco f) utio AEA- Xouif] apa f) Otto TEA Xoutfj xfj utto BEA far) eaxtv o^totwc; 8f] Beix^rjaexai, oxi xal at Otto TEB, AEA laai eiatv. 'Eav apa 8uo eui9eTai xe^tvwaiv dXXf|Xa<;, xa<; xaxa xo- pucprjv ywvtag laac, dXXrjXaii; ttoiouoiv orcep e8ei SeT^oti. If'. navxog xpiycivou (iiac xfiv TtXeupwv TtpoaexpXr] , Mar)<; f) exxoc; ywvia exaxepa<; xov evxoc xal aTtevavxiov ycoviCSv jieiCwv eaxiv. Tiaxco xpiytovov xo ABr, xal TtpoaexpepXria-dc) auxoO ^ua TtXeupa f] Br era xo A- Xeyto, oxi f) exxo<; ywvta f) utio ArA ^lei^tov eaxlv exaxepac; xwv evxoc; xal drcevavxtov iSv utto TBA, BAr yovifiv. Tex^if|a , do f] Ar 8txa xaxa xo E, xal STU^eux'deiaa f] BE expepXf]a , dw in euiSeiai; era xo Z, xal xeta-dw xrj BE far) f) EZ, xal eTteCeux'S" f) Zr, xal Sir^M V] Ar eicl xo H. 'Excel ouv iar] eaxlv f] (jiev AE xfj EI 1 , f) 8e BE xfj EZ, 8uo 8f) at AE, EB 8uat xau; TE, EZ laai etalv exaxepa exaxepcr xal ycovia rj Otto AEB ywvta xfj Otto ZEr lot] eaxiv xaxa xopucpfjv yap- pdan; apa f) AB pdaei xfj ZT iar) eaxiv, xal xo ABE xpiytovov xu ZEr xpiycovcp eaxlv laov, xal at Xoircal For let the two straight-lines AB and CD cut one an- other at the point E. I say that angle AEC is equal to (angle) DEB, and (angle) CEB to (angle) AED. A B For since the straight-line AE stands on the straight- line CD, making the angles CEA and AED, the (sum of the) angles CEA and AED is thus equal to two right- angles [Prop. 1.13]. Again, since the straight-line DE stands on the straight-line AB, making the angles AED and DEB, the (sum of the) angles AED and DEB is thus equal to two right-angles [Prop. 1.13]. But (the sum of) CEA and AED was also shown (to be) equal to two right-angles. Thus, (the sum of) CEA and AED is equal to (the sum of) AED and DEB [C.N. 1]. Let AED have been subtracted from both. Thus, the remainder CEA is equal to the remainder BED [C.N. 3]. Similarly, it can be shown that CEB and DEA are also equal. Thus, if two straight-lines cut one another then they make the vertically opposite angles equal to one another. (Which is) the very thing it was required to show. Proposition 16 For any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles. Let ABC be a triangle, and let one of its sides BC have been produced to D. I say that the external angle ACD is greater than each of the internal and opposite angles, CBA and BAG. Let the (straight-line) AC have been cut in half at (point) E [Prop. 1.10]. And BE being joined, let it have been produced in a straight-line to (point) FJ And let EE be made equal to BE [Prop. 1.3], and let EC have been joined, and let AC have been drawn through to (point) G. Therefore, since AE is equal to EC, and BE to EF, the two (straight-lines) AE, EB are equal to the two 20 ETOIXEIfiN a'. ELEMENTS BOOK 1 ycovioa xalc; Xoutau; Yioviaic; Taai rialv sxaxspa exaxepa, ucp' at; ai Taai TtXeupal Ouoxeivouaiv Tar] apa saxlv f] utto BAE xrj U7i6 ErZ. jieiCwv 8£ eaxiv f] Otto ErA xrj<; Otto ETZ- ^iriCcov apa f] Otto ATA xrj<; Otto BAE. 'O[io'ux>z Sr] xfjc; Br xex[ir]^£vr]<; Si/a Beix'Sffoexai xal f) Otto BrH, xouxsaxiv rj Otco ArA, urii^cov xal xfj? Otco ABr. Ilavxoc; apa xpiywvou ^iiac; xfiv TiXeupGv upooex- pXrj'vMarjc; f] exxog ywvia exaxepac; xfiv evxoc; xal dme- vavxiov ywviwv (jeiCuv eaxiv onep e8ei 8ric;ai. (straight-lines) Ci?, EF, respectively. Also, angle AEB is equal to angle FEC, for (they are) vertically opposite [Prop. 1.15]. Thus, the base AB is equal to the base FC, and the triangle ABE is equal to the triangle FEC, and the remaining angles subtended by the equal sides are equal to the corresponding remaining angles [Prop. 1.4]. Thus, BAE is equal to ECF. But ECD is greater than ECF. Thus, ACD is greater than BAE. Similarly, by having cut BC in half, it can be shown (that) BCG — that is to say, ACD — (is) also greater than ABC. A .F Thus, for any triangle, when one of the sides is pro- duced, the external angle is greater than each of the in- ternal and opposite angles. (Which is) the very thing it was required to show. t The implicit assumption that the point F lies in the interior of the angle ABC should be counted as an additional postulate. Ilavxoc; xpiywvou ai 860 ywviai 860 opiStov eXdaaovsc; rial Ttdvxfj ^exaXa^pavo^svai. A b r a "Eaxw xpiywvov xo ABT - Xeyco, oxi xou ABT xpiywvou ai 860 Y wv ' al §uo op^wv eXdxxovec; rial Ttdvxr] (jiexaXa^t- Pavojievai. Proposition 17 For any triangle, (the sum of) two angles taken to- gether in any (possible way) is less than two right-angles. A B CD Let ABC be a triangle. I say that (the sum of) two angles of triangle ABC taken together in any (possible way) is less than two right-angles. 21 ETOIXEIfiN a'. ELEMENTS BOOK 1 'ExpepAr](Td<j yap f\ BT iia xo A. Kal tnei xpiyiovou tou ABr exxoc; eaxi y«via r\ bnb ATA, ^ei^iov eaxi xfjc; evxoc; xai drcevavxiov xfjc; bnb ABT. xoivf] npoaxeio'dw f) bub ATB- ai dpa Otto ArA, ArB x£>v Otto ABr, BEA [leiZovez eiaiv. dXX' ai Otto ArA, ArB 860 op^dic; laai etaCv ai apa Otto ABr, BrA 860 op^wv eXdaaovec; eiaiv. o^ioicoc; 8f) 8eic;o^ev, oxi xai ai Otto BAr, ArB 860 op-dSv eXdaaovec eiai xai exi ai Otco TAB, ABT. navxoc; dpa xpiywvou ai 860 ywviai 860 op'dwv eXdac;- ovec; eiai Tidvxfj ^exaXa^pavo^evai- oTtep eBei 8eTc;ai. ir]'. navxoc; xpiycovou f] ^e[£wv TiXeupd xf]V ^ei^ova ywviav UTtoxdvei. A "Eaxw yap xpiycovov xo ABr [isiCova e)(ov xf)v Ar TtXeupdv xfjc AB- Xeyw, oxi xai ycovia f] Otto ABr (jlel^cov eaxi xfjc Otto BEA- Tkei yap [Jiei^cov eaxiv r] Ar xfjc AB, xeiai!)« xfj AB tar] f] AA, xai STieCe'JX'dw f] BA. Kai CTtei xpiyovou xoO BrA exxoc; eaxi ywvia f) Otto AAB, jieiCwv eaxi xfjc; evxoc; xai aTievavxiov xfjc; bnb ArB' iar] 8e f] Otto AAB xfj (mo ABA, ine\ xai TiXeupd f] AB xfj AA eaxiv laiy fiei^cov dpa xai f) bnb ABA xfjc; Otto ATB- ttoXXG apa f] Otto ABr ^.ei^wv eaxi xfjc; Otco ArB. navxoc; dpa xpiyovou f) [icii^wv TiXeupd xf)v ^.ei£ova ywviav UTioxeivei- ouep e8ei 8eTc;ai. Eiavxoc; xpiyiovou bnb xfjv ^ei^ova ywviav f\ ^teiC«v TiXeupd UTioxeivei. Tiaxco xpiycovov xo ABr ^leic^ova e)(ov xf]v Otco ABr yuviav xfjc; Otto BrA- Xeyw, 6x1 xai TiXeupd f] Ar TcXeupdc; xfjc; AB ^leiCwv eaxiv. For let BC have been produced to D. And since the angle ACD is external to triangle ABC, it is greater than the internal and opposite angle ABC [Prop. 1.16]. Let ACB have been added to both. Thus, the (sum of the angles) ACD and ACB is greater than the (sum of the angles) ABC and BCA. But, (the sum of) ACD and ACB is equal to two right-angles [Prop. 1.13]. Thus, (the sum of) ABC and BCA is less than two right- angles. Similarly, we can show that (the sum of) BAC and ACB is also less than two right-angles, and further (that the sum of) CAB and ABC (is less than two right- angles). Thus, for any triangle, (the sum of) two angles taken together in any (possible way) is less than two right- angles. (Which is) the very thing it was required to show. Proposition 18 In any triangle, the greater side subtends the greater angle. A For let ABC be a triangle having side AC greater than AB. I say that angle ABC is also greater than BCA. For since AC is greater than AB, let AD be made equal to AB [Prop. 1.3], and let BD have been joined. And since angle ADB is external to triangle BCD, it is greater than the internal and opposite (angle) DCB [Prop. 1.16]. But ADB (is) equal to ABD, since side AB is also equal to side AD [Prop. 1.5]. Thus, ABD is also greater than ACB. Thus, ABC is much greater than ACB. Thus, in any triangle, the greater side subtends the greater angle. (Which is) the very thing it was required to show. Proposition 19 In any triangle, the greater angle is subtended by the greater side. Let ABC be a triangle having the angle ABC greater than BCA. I say that side AC is also greater than side AB. 22 ETOIXEIfiN a'. ELEMENTS BOOK 1 Et yap \if\, fjxoi. iar) eaxlv f) Ar xfj AB fj eXdaawv tar] (lev ouv oux eaxiv f] Ar xfj AB- lor] yap dv rjv xod ytovia f] Otto ABr xfj (mo ArB- oux eaxi 8e- oux apa for) eaxlv f) Ar xfj AB. ou8e fjif|v eXdaawv eaxlv rj Ar xfj? AB- eXdaatov yap dv rjv xal ywvia f) utco ABr xfjc utio ArB- oux eaxi 8e- oux apa eXdaatov eaxlv f\ AT xfj? AB. eSsix^^ 8s, oxi oG8e Tar] eaxiv. ^ie[£c>)v apa eaxlv f] Ar xfj? AB. A B r Ilavxo? apa xpiywvou Otto xrjv (jiei^ova ywviav f] (iclCwv icXeupa uiroxeivei- oiiep e8ei 8eT?"ai. X . Ilavxo? xpiycovou ai 8uo nXeupal xfj? XoiTtfj? ^ei£ove? eiai Tidvxr] ^exaXa^3av6^evai. A "Eaxw yap xpiywvov xo ABr- Xeyw, oxi xou ABr xpiyovou ai Suo uXeupal xfjc; Xomfj? ^lei^ove? eiai Tcavxri ti£xaXa[ipav6[i£vai, ai [ie\i BA, Ar xfj? Br, ai Be AB, Br xfj? Ar, ai Be Br, TA xfj? AB. For if not, AC is certainly either equal to, or less than, AB. In fact, AC is not equal to AB. For then angle ABC would also have been equal to ACB [Prop. 1.5]. But it is not. Thus, AC is not equal to AB. Neither, indeed, is AC less than AB. For then angle ABC would also have been less than ACB [Prop. 1.18]. But it is not. Thus, AC is not less than AB. But it was shown that (AC) is not equal (to AB) either. Thus, AC is greater than AB. A B C Thus, in any triangle, the greater angle is subtended by the greater side. (Which is) the very thing it was re- quired to show. Proposition 20 In any triangle, (the sum of) two sides taken to- gether in any (possible way) is greater than the remaining (side). D B C For let ABC be a triangle. I say that in triangle ABC (the sum of) two sides taken together in any (possible way) is greater than the remaining (side). (So), (the sum of) BA and AC (is greater) than BC, (the sum of) AB 23 ETOIXEIfiN a'. ELEMENTS BOOK 1 Air))cdco yap fj BA em to A arj^eTov, xal xeiaftw xfj IA lot) f] AA, xal £Tie^£U)cd« f) Ar. 'EtcsI ouv far) eaxlv f) A A xfj Ar, ia/) eaxl xal ywvia f) uno AAr xfj Otio ATA- jiei^wv apa f) utto BrA xfjc uno AAr- xal ETxel xpiywvov eaxi xo ArB (jieiCova e)(ov xrjv utio BrA ywviav xfjc; Otto BAr, Otto 8e xfjv jiei^ova ywviav f] ^ei^wv TiXeupa Onoxeivei, f] AB apa xfjc; Br eaxi ^eic^wv. Tar) 8e r) AA xfj Ar- jiei^ovec; apa al BA, Ar xfjc Br- ojioiwc; 8f) Beii;o|jiev, oxi xal ai y.e\> AB, Br xfjc; TA [leiCovec eiaiv, ai 8e Br, FA xfjc; AB. Llavxoc; apa xpiywvou ai 860 TtXeupai xfjc; Xoiitfjc; [Lei^ovez eloi Tidvxr) uexaXa^pavo^evai- oitep e8ei Sel^ai. xa'. 'Eav xpiywvou era (iiac; xwv TtXeupwv duo xwv Tiepdxwv 860 eui£>eTai evxoc; auaxai!)waiv, ai auaxa-delaai xwv Xoitcwv xoO xpiywvou 860 TtXeupwv eXdxxovec; \ie\i eaovxai, ^tei^ova 8e ywviav 7i:epiec;ouaiv. Tpiywvou yap xoO ABr era ^iiac; xwv nXeupwv xfjc; Br dn:6 xwv Tiepdxwv xwv B, T 860 eui9eTai evxoc; auveaxdxwaav ai BA, Ar- Xeyw, oxi ai BA, Ar xwv Xoittwv xou xpiywvou 860 TtXeupwv xwv BA, Ar eXdaaovec; [lev eiaiv, [iei^ova 8s ywviav Tispisxouoi xfjv utio BAr xfjc; bub BAT. Aif|)cdw yap f] BA era xo E. xal CTtel Ttavxoc xpiywvou ai 860 TtXeupai xfjc; XoiTtfjc; (jiei^ovec; eiaiv, xoO ABE apa xpiywvou ai 860 TtXeupai ai AB, AE xfjc; BE ^tei^ovec eiaiv xoivf) Ttpoaxeia'dw f) EE ai apa BA, Ar xwv BE, Er jiei^ovec; eiaiv. TtdXiv, CTtel xou TEA xpiywvou ai 860 TtXeupai ai TE, EA xfjc; EA [iciCovec; eiaiv, xoivf] Ttpoaxeia'dw f] AB- ai EE, EB apa xwv EA, AB jieiCovec; eiaiv. dXXa xwv BE, Er [leiCovec; e8ei)edr)aav ai BA, Ar- TtoXXw apa ai BA, Ar xwv BA, Ar ^d^ovec; eiaiv. ndXiv, eitei Ttavxoc; xpiywvou f) exxoc; ywvia xfjc; evxoc; xal aTtevavxiov jieii^wv eaxiv, xou EAE apa xpiywvou f) exxoc; ywvia f) bub BAT ^lei^wv eaxi xfjc; utio TEA. 81a xauxa xoivuv xal xou ABE xpiywvou f] exxoc; ywvia f) bub and BC than AC, and (the sum of) BC and CM than AB. For let have been drawn through to point D, and let AD be made equal to CA [Prop. 1.3], and let DC have been joined. Therefore, since DA is equal to AC, the angle ADC is also equal to ACD [Prop. 1.5]. Thus, BCD is greater than ADC. And since DCB is a triangle having the angle BCD greater than BDC, and the greater angle subtends the greater side [Prop. 1.19], DB is thus greater than BC. But DA is equal to AC. Thus, (the sum of) BA and AC is greater than BC. Similarly, we can show that (the sum of) AB and BC is also greater than CA, and (the sum of) and C.4 than AB. Thus, in any triangle, (the sum of) two sides taken to- gether in any (possible way) is greater than the remaining (side). (Which is) the very thing it was required to show. Proposition 21 If two internal straight-lines are constructed on one of the sides of a triangle, from its ends, the constructed (straight-lines) will be less than the two remaining sides of the triangle, but will encompass a greater angle. For let the two internal straight-lines BD and DC have been constructed on one of the sides BC of the tri- angle ABC, from its ends B and C (respectively) . I say that BD and DC are less than the (sum of the) two re- maining sides of the triangle BA and AC, but encompass an angle BDC greater than BAG. For let BD have been drawn through to E. And since in any triangle (the sum of any) two sides is greater than the remaining (side) [Prop. 1.20], in triangle ABE the (sum of the) two sides AB and AE is thus greater than BE. Let EC have been added to both. Thus, (the sum of) BA and AC is greater than (the sum of) BE and EC. Again, since in triangle CED the (sum of the) two sides CE and ED is greater than CD, let DB have been added to both. Thus, (the sum of) CE and EB is greater than (the sum of) CD and DB. But, (the sum of) BA and AC was shown (to be) greater than (the sum of) BE and EC. Thus, (the sum of) BA and AC is much greater than 24 ETOIXEIfiN a'. ELEMENTS BOOK 1 TEB (jiei^wv iav. tfj? Otto BAI\ dXXd xfjc; uno TEB (jiei^wv sSeix'&T) T) utto BAr- ttoXXG apa f] uno BAr ^lei^wv eaxi xfjc Oko BAr. 'Edv apa xpiyGvou era [iidc; xGv TtXeupGv dno xSv Ttepdxcov Buo eu-delai evxo<; auaxa'dwaiv, ai auaxa'deTaai x£>v XoitcGv xou xpiyGvou Buo TtXeupGv eXdxxovec; jiev eiaiv, ^iei£ova Be ywviav nepiexouaiv oitep eBei SeT^ai. (the sum of) BD and DC. Again, since in any triangle the external angle is greater than the internal and opposite (angles) [Prop. 1.16], in triangle CDE the external angle BDC is thus greater than CED. Accordingly, for the same (reason), the external angle CEB of the triangle ABE is also greater than BAC. But, BDC was shown (to be) greater than CEB. Thus, BDC is much greater than BAC. Thus, if two internal straight-lines are constructed on one of the sides of a triangle, from its ends, the con- structed (straight-lines) are less than the two remain- ing sides of the triangle, but encompass a greater angle. (Which is) the very thing it was required to show. x(3'. 'Ex xpiGv eu'deiwv, off eiaiv Xoou. xpioi xau; So'deiaau; [eOifteiau;] , xpiyovov auox^aaCTdai - Set Be xd<; 860 xfjc Xomfjc (jidi^ovac; eTvai Ttdvxy] (iexaXa^pavo^eva<; [Bid xo xai navxoc xpiyovou xac; Buo rcXeupac xfjc Xomfjt; [iei^ovac eTvai Tidvxr] (i£xaXa^i[3avou£vac;] . Proposition 22 To construct a triangle from three straight-lines which are equal to three given [straight-lines]. It is necessary for (the sum of) two (of the straight-lines) taken together in any (possible way) to be greater than the remaining (one), [on account of the (fact that) in any triangle (the sum of) two sides taken together in any (possible way) is greater than the remaining (one) [Prop. 1.20] ]. A B C z H e 'Eoxwaav ai Bo-deTaai xpeTc; euiMai ai A, B, T, Sv ai Buo xfjc Xomfjc \LZiC,ovec, eoxwoav Kavxr) (lexaXa^pavo^ievai, ai ]izv A, B xfj<; T, ai Be A, T xrj<; B, xal exi ai B, T xfjc A- Bel Br) ex xGv lawv xdic; A, B, T xpiywvov auaxriaacrdai. 'Exxeicrdco xic; eMela f] AE KCTiepaajievr) [lev xaxd xo A aTteipoc; Be xaxd xo E, xai xeicrdco xfj [ism A lot] f] AZ, xfj Be B tar) f] ZH, xfj Be T Tar) rj HO- xai xevxptp jiev xG Z, Biaaxrj^iaxi Be iS ZA xuxXo<; yeypdcpTdw 6 AKA - TtdXiv xevxpcp |iev xG H, Biaaxr^iaxi Be xG H6 xuxXoc; yeypdcp-dco 6 KA0, xai ejie^eu)cd«aav ai KZ, KH- Xeyco, oxi ex xpiGv eMeiGv xGv lacov xoii<; A, B, T xpiywvov ouveaxaxai xo KZH. Tkei yap xo Z arjueTov xevxpov eaxi xou AKA xuxXou, iar] eaxiv f) ZA xfj ZK- dXXd f] ZA xfj A eaxiv lar\. xai f) D F G H Let A, B, and C be the three given straight-lines, of which let (the sum of) two taken together in any (possible way) be greater than the remaining (one). (Thus), (the sum of) A and B (is greater) than C, (the sum of) A and C than £?, and also (the sum of) B and C than A. So it is required to construct a triangle from (straight-lines) equal to A, B, and C. Let some straight-line DE be set out, terminated at D, and infinite in the direction of E. And let DF made equal to A, and FG equal to 5, and GH equal to C [Prop. 1.3]. And let the circle DKL have been drawn with center F and radius FD. Again, let the circle KLH have been drawn with center G and radius GH. And let KF and KG have been joined. I say that the triangle KFG has 25 ETOIXEIfiN a'. ELEMENTS BOOK 1 KZ apa xfj A eaxiv Xar\. TidXiv, ETtd to H g/^eTov xevxpov eaxl xoO AK0 xuxXou, lor] eaxlv #] H9 xfj HK- dXXa f] H9 xfj T saxiv laiy xal f\ KH apa xfj T eaxiv lor), eaxl Be xal f] ZH xfj B tar) - ai xpdc; apa eCWteTai ai KZ, ZH, HK xpial xaic A, B, r i'aai daiv. 'Ex xpiGv apa eu-daGv xGv KZ, ZH, HK, ou. daiv laai xpial xaic; BoiJeiaaic eCWteiau; xau; A, B, T, xpiycovov auveaxaxai xo KZH oicep e§ei itoifjaai. been constructed from three straight-lines equal to A, B, and C. For since point F is the center of the circle DKL, FD is equal to FK. But, FD is equal to A. Thus, KF is also equal to A. Again, since point G is the center of the circle LKH, GH is equal to GK. But, GH is equal to C. Thus, KG is also equal to C. And FG is also equal to B. Thus, the three straight-lines KF, FG, and GA are equal to A, B, and C (respectively). Thus, the triangle KFG has been constructed from the three straight-lines KF, FG, and GK, which are equal to the three given straight-lines A, B, and C (re- spectively) . (Which is) the very thing it was required to do. xy . npoc; xfj Bo-deia/] eO'dda xal xG Ttpoc; auxfj or\[Lsicd xfj 5oi9dar] yovia eO'duypa^.u.to I'arjv ywviav EU'duypa^ov auaxrpacrdai. Proposition 23 To construct a rectilinear angle equal to a given recti- linear angle at a (given) point on a given straight-line. z H B "Eaxco f\ \ie\> Bcdeiaa eO'dda fj AB, xo Be Ttpoc; auxfj arjueTov xo A, f] 8e 8o$daa ywvia eO'duypa^oc; f] Oico ArE- 8eT 6f] TCpoc; xfj Scdeiar) eui&eia xfj AB xal xG npoc, auxfj ar)udcp xG A xfj 8oi9e[ar] ywvia eG^uypd^iJiip xfj uito ArE 'larjv ycoviav einJuypaijijj.ov auaxfpacrdai. EiX^cpdo ecp' exaxepac; xGv TA, TE xuxovxa arj^ida xa A, E, xal eice^eux'dco f) AE- xal ex xpiGv eui5eiGv, ai eiaiv law. xpial xaic; IA, AE, TE, xpiycovov auveaxdxco xo AZH, Gaxe larjv elvai x/jv [lev TA xfj AZ, xf]v 8s TE xfj AH, xal exi xf)v AE xfj ZH. 'End ouv Buo ai AT, TE 86o xaic; ZA, AH I'aai dalv exaxepa exaxepa, xal pdau; f] AE pdaei xfj ZH lot], yovia dpa f] utco ArE ywvia xfj utco ZAH eaxiv iarj. npoc; dpa xfj Bo'deiar] eu-deia xfj AB xal xG npoc; auxfj arjueiw xG A xfj Bo'deiar] ywvia eu'duypd^w xfj utco ArE Xar] yovia euiJuypa^oc; auveaxaxai f) utco ZAH- oTtep eBei Ttoirjaai. F G B Let AB be the given straight-line, A the (given) point on it, and DCE the given rectilinear angle. So it is re- quired to construct a rectilinear angle equal to the given rectilinear angle DCE at the (given) point A on the given straight-line AB. Let the points D and E have been taken at random on each of the (straight-lines) CD and CE (respectively), and let DE have been joined. And let the triangle AFC have been constructed from three straight-lines which are equal to CD, DE, and CE, such that CD is equal to AF, CE to AG, and further DE to FG [Prop. 1.22]. Therefore, since the two (straight-lines) DC, CE are equal to the two (straight-lines) FA, AG, respectively, and the base DE is equal to the base FG, the angle DCE is thus equal to the angle FAG [Prop. 1.8]. Thus, the rectilinear angle FAG, equal to the given rectilinear angle DCE, has been constructed at the (given) point A on the given straight-line AB. (Which 2G ETOIXEIfiN a'. ELEMENTS BOOK 1 x5'. 'Edv Buo xpiytova xdc; Buo nXeupdc; [xdic;] 860 TtXeupaTc; I'aac; eyr\ exaxepav exaxepa, xrjv Be ytovlav xfjc; ytovlac; ^e[£ova ex?) xr ] v utto T " v i° wv eu-deiwv Tiepie)(o^evr]v, xal xf]v pdaiv Tfj? pdaetoc; ^lel^ova ec;ei. A A B \ \/7 "Eaxco Buo xpiywva xd ABr, AEZ xdc; Buo TtXeupdc; xdc; AB, Ar xaic; Buo TtXeupaTc; xaic; AE, AZ I'aac; exovxa exaxepav exaxepa, xrjv |iev AB xfj AE xf]v Be Ar xfj AZ, f) Be Ttpoc; xa> A ycovia xfjc; Tipoc; xai A ycoviac; ^xei^cov eaxw Xeyco, 6x1 xal pdaig f] Br pdaewc; xfjc; EZ ^ei^wv eaxiv. Tkel y«P ^ei^wv f) Otto BAr ywvia xfjc; utio EAZ ycoviac;, auveaxdxw Ttpoc; xrj AE eu-deia xal x65 Tipoc; auxfj a/jjieicp xw A xfj Otto BAr ycovia Xat] i\ utto EAH, xal xeiai9« OTtoxepa xwv Ar, AZ lay] f) AH, xal CTie^eu)(i9«oav ai EH, ZH. Tkel ouv Xat] eaxlv f] [Lev AB xfj AE, f] Be Ar xrj AH, Buo Br) ai BA, Ar Buai xaic; EA, AH i'aai eialv exaxepa exaxepa- xal yiovia f] utio BAr ywvia ^fi utio EAH Xai]- pdaic; apa f) Br pdaei xrj EH eaxiv Tar). TidXiv, CTiel Xat] eaxlv f| AZ xfj AH, Tar) eaxl xal f] Otto AHZ ycovia xfj utio AZH' [ici^wv apa f) utio AZH xfjc utio EHZ' TioXXai apa [ici^iov eaxlv f) utio EZH xfjc; Otto EHZ. xal ercel xpiycovov eaxi xo EZH jiet^ova e/ov xf]v utio EZH ywviav xfjc; utio EHZ, utio Be xrjv jiet^ova ywviav f) [leii^v TtXeupd UTraxeivei, ^eiCwv apa xal TtXeupd i\ EH xfjc; EZ. Xor] Be f) EH xfj BE ^ei^tov apa xal f) Br xfjc; EZ. 'Eav apa 860 xpiywva xdc; Buo TtXeupdc; Bual TtXeupaTc; laac; syr\ exaxepav exaxepa, xf)v Be ywvlav xfjc; ywviac; ^eii^ova exT) xfjv utco xGv tawv eu^eiOv Ttepiexo^ievr]v, xal xf)v pdaiv xfjc; pdaewc; ^tei^ova e^er oTtep eBei Bel^ai. is) the very thing it was required to do. Proposition 24 If two triangles have two sides equal to two sides, re- spectively, but (one) has the angle encompassed by the equal straight-lines greater than the (corresponding) an- gle (in the other), then (the former triangle) will also have a base greater than the base (of the latter) . A D Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF, respectively. (That is), AB (equal) to DE, and AC to DF. Let them also have the angle at A greater than the angle at D. I say that the base BC is also greater than the base EF. For since angle BAC is greater than angle EDF, let (angle) EDG, equal to angle BAC, have been constructed at the point D on the straight-line DE [Prop. 1.23]. And let DC be made equal to either of AC or DF [Prop. 1.3], and let EG and FG have been joined. Therefore, since AB is equal to DE and AC to DC, the two (straight-lines) BA, AC are equal to the two (straight-lines) ED, DG, respectively. Also the angle BAC is equal to the angle EDG. Thus, the base BC is equal to the base EG [Prop. 1.4]. Again, since DF is equal to DG, angle DGF is also equal to angle DFG [Prop. 1.5]. Thus, DFG (is) greater than EGF . Thus, EFG is much greater than EGF. And since triangle EFG has angle EFG greater than EGF, and the greater angle is subtended by the greater side [Prop. 1.19], side EG (is) thus also greater than EF. But EG (is) equal to BC. Thus, BC (is) also greater than EF. Thus, if two triangles have two sides equal to two sides, respectively, but (one) has the angle encompassed by the equal straight-lines greater than the (correspond- ing) angle (in the other), then (the former triangle) will also have a base greater than the base (of the latter). 27 ETOIXEIfiN a'. ELEMENTS BOOK 1 xz. 'Eav 860 xpiywva xdc 860 TtXeupdc. 8uol TtXeupdlc laac zyj\ exaxepav exaxepa, xf)v 8e pdaiv xfjc. pdaewc. ^.ei^ova e/r), xal x/)v yioviav xfjc ycoviac. ^ei^ova ei;ei xf]v utto xfiv i'ouv eui)eifiv Ttepiexojievrjv. A 'Eaxco 860 xpiywva xd ABr, AEZ xdc. 860 TtXeupdc. xdc. AB, Ar xalc 860 TtXeupaTc xalc AE, AZ I'aac. exovxa exaxepav exaxepa, xrjv jiev AB xfj AE, xf|v 6e Ar xfj AZ- pdaic. 8e f] BT pdaeoc xfj? EZ ue(C«v eaxw Xey«, oxi xal ycovia f] utio BAr ycoviac xfjc utio EAZ jiei^wv eaxiv. EE yap [ir|, fjxoi Tar) eaxiv auxfj rj eXdaatov Tar) ^tev ouv oux eaxiv f] utio BAT xfj Otto EAZ- lay) yap av rjv xal pdaic f] Br pdaei xfj EZ- oux eaxi 8e. oux apa Tar) eaxi ycovia f] utio BAr xrj utio EAZ- ouSe |if]v eXdaawv eaxiv f] utio BAr xfjc utio EAZ- eXdaatov yap av rjv xal pdaic. f] Br pdaeGJc. xfjc EZ- oux ecru 6e- oux apa eXdaawv eaxiv f) utio BAr ycovia xfjc. utio EAZ. eSeL/iS/] 8e, 6x1 ouSe Tar)- ^eii^tov apa eaxiv f| utio BAr xfjc utio EAZ. 'Eav apa 860 xpiycova xdc. 860 TiXeupdc. 8ual TrXeupdic. 1'aac. exfj exaxepav exaxepa, xf]v 8e Paaiv xfjc. pdaetoc [ici^ova sxTli Tr ] v ywviav xfjc, yioviac. jiei^ova e^ei xfjv uno xcov 'lacov eui5eifiv Tiepie/o(jievr]V- oiiep e8ei SeT^ai. Xf'. 'Eav Suo xpiyova xdc. 860 ycoviac. Sual ycoviaic. 1'aac. 'iyjl exaxepav exaxepa xal puav TtXeupdv ^iia TiXeupa iar]v fjxoi xf)V Tipoc. xalc laaic ycoviaic fj xfjv UTtoxeivouaav utio ^uav iSv Tacov ytoviaiv, xal xdc XoiTidc. TiXeupdc. xalc; XoiTidic. TiXeupdic. 1'aac. e5ei [exaxepav exaxepa] xal xfjv XoiTifjv yioviav xfj XoiTifj ycovia. TSaxco 860 xpiycova xd ABr, AEZ xdc. 8uo ycoviac. xdc (Which is) the very thing it was required to show. Proposition 25 If two triangles have two sides equal to two sides, respectively, but (one) has a base greater than the base (of the other), then (the former triangle) will also have the angle encompassed by the equal straight-lines greater than the (corresponding) angle (in the latter) . A E F Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF, respectively (That is), AB (equal) to DE, and AC to DF. And let the base BC be greater than the base EF. I say that angle BAG is also greater than EDF. For if not, (BAC) is certainly either equal to, or less than, (EDF). In fact, BAC is not equal to EDF. For then the base BC would also have been equal to the base EF [Prop. 1.4]. But it is not. Thus, angle BAC is not equal to EDF . Neither, indeed, is BAC less than EDF . For then the base BC would also have been less than the base EF [Prop. 1.24]. But it is not. Thus, angle BAC is not less than EDF. But it was shown that (BAC is) not equal (to EDF) either. Thus, BAC is greater than EDF . Thus, if two triangles have two sides equal to two sides, respectively, but (one) has a base greater than the base (of the other), then (the former triangle) will also have the angle encompassed by the equal straight-lines greater than the (corresponding) angle (in the latter). (Which is) the very thing it was required to show. Proposition 26 If two triangles have two angles equal to two angles, respectively, and one side equal to one side — in fact, ei- ther that by the equal angles, or that subtending one of the equal angles — then (the triangles) will also have the remaining sides equal to the [corresponding] remaining sides, and the remaining angle (equal) to the remaining angle. 28 ETOIXEIfiN a'. ELEMENTS BOOK 1 0x6 ABr, BrA 5ual xau; 0tt:6 AEZ, EZA laac; eyrovxa exaxepav exaxepa, xf)v ^tev 0x6 ABr xfj 0x6 AEZ, xf)v 8s Otio BEA xfj 0x6 EZA- exexco Se xal ^iav TtXeupdv [iia xXeupa larjv, Ttpoxepov xfjv upoc; xalc; laai? ytavtaic; xf)v Br xfj EZ- Xeyco, oxi xal xd<; Xoixac; xXeupac; xalc; Xoixdic; xXeupau; laa? ec;ei exaxepav exaxepa, xf)v ^tev AB xfj AE xf]v 8s Ar xfj AZ, xal xf)V Xoixf)v ycovlav xfj XoiTtfj ycovla, xfjv 0x6 BAr xfj 0x6 EAZ. El yap dviaoc; eaxiv f) AB xfj AE, aOxcov ^telCcov eaxiv. eaxco ^el^cov f) AB, xal xeladco xfj AE lot) f) BH, xal exci^eu^-dco f] HT. 'EtceI ouvl'or) eoxlv f) jiev BH xfj AE, f) 8s Br xfj EZ, 660 8f) al BH, Br 6ual xalc; AE, EZ I'oai eiolv exaxepa exaxepa- xal ycovla f] 0x6 HBT ycovla xfj 0x6 AEZ I'or) eoxlv pdaic; apa f) Hr pdaei xfj AZ I'or) eaxiv, xal xo HBr xplycovov xco AEZ xpiycovcp Taov eoxlv, xal al Xoixal ycovlai xau; Xoixdic; ycovlaic; Taai eoovxai, Ocp' ac; al Taai TiXeupal OTtoxelvouoiv Tar) apa f) Otto HrB ycovla xfj 0x6 AZE. dXXa f) 0x6 AZE xfj Otto BEA UTtoxeixai Tar)' xal f) 0x6 BrH apa xfj Otto BEA Tar) eoxlv, f) eXdaocov xfj jiel^ovi- oxep dBuvaxov. oOx dpa dviaoc; eaxiv f) AB xfj AE. I'or) dpa. eaxi 8e xal f) Br xfj EZ Tar)' 860 Sf) al AB, Br 8uol xalc; AE, EZ Taai eiolv exaxepa exaxepa' xal ycovla f) 0x6 ABr ycovla xfj 0x6 AEZ eaxiv Tar)' pdoig apa f) Ar pdaei xfj AZ I'or) eaxiv, xal Xoixf) ycovla f) 0x6 BAr xfj Xoixfj ycovla xfj O716 EAZ I'or) eoxlv. AXXd Sf) TidXiv eaxcoaav al Otto xdc; Taac; ycovlac; xXeupal Oxoxelvouaai I'oai, cbc; f) AB xfj AE' Xeyco xdXiv, oxi xal al Xomal xXeupal xalc; Xoixdic; TiXeupdic; Taai eoovxai, f) (lev Ar xfj AZ, f) 8e Br xfj EZ xal exi f) Xomf) ycovla i\ 0x6 BAr xfj XoiTtfj ycovla xfj Otto EAZ Tar) eaxiv. Ei yap dviooc; eaxiv f) Br xfj EZ, [ila aOxcov jiel^cov eaxiv. eaxco ^elc^cov, ei 8uvax6v, f) Br, xal xeladco xfj EZ Tar) f) B9, xal CTie^e0)(i9co fj A6. xal ctcci Tor) eoxlv f) ^iev B9 xfj EZ f) 8e AB xfj AE, 660 Sf) al AB, B6 8ual xalc; AE, EZ Taai eiolv exaxepa exapepa- xal ycovlac; Taac; xepiexouaiv pdoic; apa f] AO pdaei xfj AZ Tar) eoxlv, xal xo AB9 xplycovov xco AEZ xpiycovcp Taov eoxlv, xal al Xoixal ycovlai xalc; Xoixdic; ycovlaic; I'oai eoovxai, Ocp' ac; al Taac; TtXeupal O-xoxelvouaiv lor) apa eaxiv f) 0x6 B0A ycovla xfj 0x6 EZA. dXXa f) Otto Let ABC and DEF be two triangles having the two angles ABC and BCA equal to the two (angles) DEF and EFD, respectively. (That is) ABC (equal) to DEF , and BCA to EFD. And let them also have one side equal to one side. First of all, the (side) by the equal angles. (That is) BC (equal) to EF. I say that they will have the remaining sides equal to the corresponding remain- ing sides. (That is) AB (equal) to DE, and AC to DF. And (they will have) the remaining angle (equal) to the remaining angle. (That is) BAC (equal) to EDF. D For if AB is unequal to DE then one of them is greater. Let AB be greater, and let BG be made equal to DE [Prop. 1.3], and let GC have been joined. Therefore, since BG is equal to DE, and BC to EF, the two (straight-lines) GB, BC^ are equal to the two (straight-lines) DE, EF, respectively. And angle GBC is equal to angle DEF. Thus, the base GC is equal to the base DF, and triangle GBC is equal to triangle DEF, and the remaining angles subtended by the equal sides will be equal to the (corresponding) remaining angles [Prop. 1.4]. Thus, GCB (is equal) to DFE. But, DFE was assumed (to be) equal to BCA. Thus, BCG is also equal to BCA, the lesser to the greater. The very thing (is) impossible. Thus, AB is not unequal to DE. Thus, (it is) equal. And BC is also equal to EF. So the two (straight-lines) AB, BC are equal to the two (straight- lines) DE, EF, respectively. And angle ABC is equal to angle DEF. Thus, the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF [Prop. 1.4]. But, again, let the sides subtending the equal angles be equal: for instance, (let) AB (be equal) to DE. Again, I say that the remaining sides will be equal to the remain- ing sides. (That is) AC (equal) to DF, and BC to EF. Furthermore, the remaining angle BAC is equal to the remaining angle EDF. For if BC is unequal to EF then one of them is greater. If possible, let BC be greater. And let BH be made equal to EF [Prop. 1.3], and let AH have been joined. And since BH is equal to EF, and AB to DE, the two (straight-lines) AB, BH are equal to the two 29 ETOIXEIfiN a'. ELEMENTS BOOK 1 EZA xfj utio BEA eaxiv Tar) - xpiywvou 8f] xou AGE f] exxoc ywvia f] Otto BOA iar) eaxi xfj evxoc; xai dracvavxiov xfj Otto BEA- oitep dSuvaxov. oux dpa dviaoc; eaxiv f\ BE xfj EZ- lor\ dpa. eaxl Se xai f] AB xfj AE iar). 8uo Sf] ai AB, BE 8uo xdic; AE, EZ laai eialv exaxepa exaxepa- xai yoviac; laac; Tiepiexouar Pdaic; dpa f) Ar pdaei xfj AZ iar) eaxiv, xai xo ABE xpiytovov iu AEZ xpiywvtp laov xai Xomf] yovia f] utio BAE xfj Xoircf] yovia xfj utio EAZ far). 'Edv dpa 860 xpiycova xac Suo ywvtac; 8uoi yoviaic; laac exaxepav exaxepa xai ^tiav TiXeupav [j.ia TiXeupd i'arjv fjxoi xfjv Ttpoc; xdic; Taaic; ycoviaic;, *) xr)V UTioxeivouaav Otto (jiiav xwv Tacov ywviwv, xai xac; Xomdc; nXeupag xdic; XoiTtdic TtXeupdic; laac; e$ei xai xf|V Xomf]V ytoviav xfj Xomfj ycoviqr oTiep e8ei 8eTc;ai. t The Greek text has "BG, BC", which is obviously a mistake. 'Edv etc; 860 eu-fJeiac; eu'de'ia epiraTtxouaa xdc; evaXXac; ywvia; laa; dXXiqXai; Ttoifj, TiapdXXrjXoi eaovxai dXXr]Xai; ai eMelai. H El? yap 860 eMeia; xac AB, EA eui5ela e^iraTixouaa f] EZ xdc; evaXXd? ycovia; xdc; utio AEZ, EZA i'aa; dXXrjXai; Ttoieixoy Xeyw, oxi TtapdXXrjXo; eaxiv f) AB xfj EA. EE yap [if], ex(3aXX6^£vai ai AB, EA au^Tieaouvxai rjxoi E7xi xd B, A |iepr) fj era xd A, E. expepXiqaOwaav xai ouji- rajtxexcoaav era xd B, A (iepr) xaxd xo H. xpiywvou 8f) xou HEZ f) exxo; ywvia f] utio AEZ iar) eaxi xfj evxoc; xai arce- vavxiov xfj utio EZH- oTtep eaxiv dSuvaxov oux dpa ai AB, AE expaXXojievai au^TteaoOvxai era xd B, A [iepr]. ojioiio; (straight-lines) DE, EF, respectively. And the angles they encompass (are also equal). Thus, the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF, and the remaining angles subtended by the equal sides will be equal to the (corresponding) remaining angles [Prop. 1.4]. Thus, angle BHA is equal to EFD. But, EFD is equal to BCA. So, in triangle AHC, the external angle BHA is equal to the internal and opposite angle BCA. The very thing (is) impossi- ble [Prop. 1.16]. Thus, BC is not unequal to EF. Thus, (it is) equal. And AB is also equal to DE. So the two (straight-lines) AB, BC are equal to the two (straight- lines) DE, EF, respectively. And they encompass equal angles. Thus, the base AC is equal to the base DF, and triangle ABC (is) equal to triangle DEF, and the re- maining angle BAG (is) equal to the remaining angle EDF [Prop. 1.4]. Thus, if two triangles have two angles equal to two angles, respectively, and one side equal to one side — in fact, either that by the equal angles, or that subtending one of the equal angles — then (the triangles) will also have the remaining sides equal to the (corresponding) re- maining sides, and the remaining angle (equal) to the re- maining angle. (Which is) the very thing it was required to show. Proposition 27 If a straight-line falling across two straight-lines makes the alternate angles equal to one another then the (two) straight-lines will be parallel to one another. G For let the straight-line EF, falling across the two straight-lines AB and CD, make the alternate angles AEF and EFD equal to one another. I say that AB and CD are parallel. For if not, being produced, AB and CD will certainly meet together: either in the direction of B and D, or (in the direction) of A and C [Def. 1.23]. Let them have been produced, and let them meet together in the di- rection of B and D at (point) G. So, for the triangle 30 ETOIXEIfiN a'. ELEMENTS BOOK 1 8f) 8ei)fdf|aexai, oxi ou8e em xd A, E at 8s era (ir]8exepa xd ^tepr] aujiTUTixouaai TcapdXXrjXot etmv TiapdXXr)Xoc; apa eaxiv #1 AB xrj rA. 'Edv apa etc; 8uo eu'deiac; eu'dela ejiTUTCxouaa xdc; evaXXdi; ywvtac; faac; dXXfjXaic; Tioifj, TiapdXXr]Xoi eaovxai at euiSelai- oTiep e8ei 8eTc;ai. XT]'. 'Edv etc; 8uo ei/deiac; eO'deia ejirarcxouaa xf|v exxoc; ywvtav xrj evxog xal dnevavxtov xat era xd auxa \iipt) far)v Tioifj fj xdc; evxoc; xal era xd auxa [ispr] Sualv opdal? faac, TiapdXXrjXoi Eaovxai dXXfjXaic; at eO'delai. Etc; yap 60o ei/deiac; xa<; AB, IA eui5eTa ejiraTcxouaa f] EZ xf|v exxoc; ycoviav xf]v utco EHB xfj evxoc; xal dnevavxbv yovta xfj utio HOA \'ar)v tcoicixgj fj xdc; evxoc; xat stil xd auxa [iepi] xdc; utio BH0, H0A §uatv op'dau; faac; - Xeyto, oxi TiapdXXr)X6<; eaxiv f] AB xrj TA. 'EticI yap far] eaxiv f) utio EHB xrj utco HOA, dXXd f] utco EHB xrj Otto AH0 eaxiv far), xal f) utio AH0 apa xrj utio H6A eaxiv far) - xat etaiv evaXXd^' TcapdXXr]Xoc; apa eaxtv f) AB xrj TA. ndXiv, eitet at utco BH0, H8A 8uo op'dalc; taai etaiv, etat 8e xat at utco AH8, BH0 8uatv opiate; t'aai, at apa utco AH0, BH0 xalc; utco BH0, H0A taai etaiv xoivf] dcpr]pf)adw f] utio BH6 - Xomf] apa f] utco AH0 XoiTcfj xrj utco HOA eaxiv far) - xat eiaiv evaXXd^ - TcapdXXrjXoc apa eaxiv f) AB xrj TA. 'Edv apa etc; Suo eu'deiac; eu'deta ejiTciTcxouaa xf)v exxoc; ycoviav xrj evxoc; xat aTcevavxtov xat era xd auxa [ispt] farjv GEF, the external angle AEF is equal to the interior and opposite (angle) EFG. The very thing is impossible [Prop. 1.16]. Thus, being produced, AB and CD will not meet together in the direction of B and D. Similarly, it can be shown that neither (will they meet together) in (the direction of) A and C. But (straight-lines) meeting in neither direction are parallel [Def. 1.23]. Thus, AB and CD are parallel. Thus, if a straight-line falling across two straight-lines makes the alternate angles equal to one another then the (two) straight-lines will be parallel (to one another) . (Which is) the very thing it was required to show. Proposition 28 If a straight-line falling across two straight-lines makes the external angle equal to the internal and oppo- site angle on the same side, or (makes) the (sum of the) internal (angles) on the same side equal to two right- angles, then the (two) straight-lines will be parallel to one another. For let EF, falling across the two straight-lines AB and CD, make the external angle EGB equal to the in- ternal and opposite angle GHD, or the (sum of the) in- ternal (angles) on the same side, BGH and GHD, equal to two right-angles. I say that AB is parallel to CD. For since (in the first case) EGB is equal to GHD, but EGB is equal to AGH [Prop. 1.15], AGH is thus also equal to GHD. And they are alternate (angles). Thus, AB is parallel to CD [Prop. 1.27] . Again, since (in the second case, the sum of) BGH and GHD is equal to two right-angles, and (the sum of) AGH and BGH is also equal to two right-angles [Prop. 1.13], (the sum of) AGH and BGH is thus equal to (the sum of) BGH and GHD. Let BGH have been subtracted from both. Thus, the remainder AGH is equal to the remainder GHD. And they are alternate (angles). Thus, AB is parallel to CD [Prop. 1.27]. 31 ETOIXEIfiN a'. ELEMENTS BOOK 1 Ttoifj fj xd? evxo? xal era xd auxd [iepr) 8ualv op'ddi? Taa?, TiapdXXr)Xoi eaovxai otl eu'dslai- oracp s8ei Bsi^ai. X0'. H d? xag TiapaXXf|Xou? euifteia? sMeia ejiraTixouaa xd? te evaXXdc; ywvia? taa? dXXrjXai? Traiei xal xfjv sxxo? xfj svxo? xal dtTievavTiov larjv xal xd? evxo? xal era xa auxd y.spf] Sualv op'ddi? Taa?. El? yap TiapaXXf|Xou? eui9da? xd? AB, EA su-deia s^tutitetco f] EZ- Xsyco, oxi xd? evaXXd?" ytovia? xd? Otto AH9, H9A Taa? Tioiei xal xf|v exxo? yioviav xf)v Otto EHB XT] evxo? xal dmsvavxbv xfj Otto H9A Tar]v xal xa? svxo? xal stxI xd auxd [iepf] xd? utio BH6, H9A Sualv opi&al? Taa?. EE yap aviao? saxiv f) utio AH9 xfj utio H9A, \i{<x auxcov iiei^wv saxiv. Eaxco [lei^cov f) utio AH9- xoivf] Tipooxeiai9« f] utio BH9- ai apa utio AH9, BH9 xwv Otto BH9, H9A [idZoMSQ eiaiv. dXXd at utio AH9, BH9 Sualv op'ddi? Taai slow, [xal] ai apa utio BH9, H9A Suo 6p$Gv sXdaaove? riaiv. ai §£ dm' eXaaaovwv f] Suo opiSwv expaXXo^ievai si? aTisipov aupuibixouaiv ai apa AB, EA ixpaXXo^ievai si? draipov au^Tisaouvxai- ou au^iraTixouai Se Bid xo Tia- paXXf|Xou? auxd? OiioxeTadai- oux apa dviao? eaxiv f) Otto AH9 xrj utio H9A- i'ar) apa. dXXd f] utio AH9 xfj Otto EHB eaxiv Tary xal f] utio EHB apa xfj Otto H9A eaxiv tar) - xoivf] Tipoaxeia'do f) Otto BH9" ai apa utio EHB, BH9 xal? utio BH9, H9A Taai eiaiv. dXXd ai utio EHB, BH9 Suo opflal? Taai eiaiv xal ai Otto BH9, H9A apa Suo op'ddi? Taai eiaiv. C H apa ei? xd? TtapaXXf|Xou? eOiJeia? eu'dela e^traTixouaa xd? xe evaXXd?" ywvia? Taa? dXXf]Xai? TioieT xal xfjv exxo? xfj evxo? xal aTievavxiov Tarjv xal xd? evxo? xal era xd auxd Thus, if a straight-line falling across two straight-lines makes the external angle equal to the internal and oppo- site angle on the same side, or (makes) the (sum of the) internal (angles) on the same side equal to two right- angles, then the (two) straight-lines will be parallel (to one another). (Which is) the very thing it was required to show. Proposition 29 A straight-line falling across parallel straight-lines makes the alternate angles equal to one another, the ex- ternal (angle) equal to the internal and opposite (angle), and the (sum of the) internal (angles) on the same side equal to two right-angles. For let the straight-line EF fall across the parallel straight-lines AB and CD. I say that it makes the alter- nate angles, AGH and GHD, equal, the external angle EGB equal to the internal and opposite (angle) GHD, and the (sum of the) internal (angles) on the same side, BGH and GHD, equal to two right-angles. For if AGH is unequal to GHD then one of them is greater. Let AGH be greater. Let BGH have been added to both. Thus, (the sum of) AGH and BGH is greater than (the sum of) BGH and GHD. But, (the sum of) AGH and BGH is equal to two right-angles [Prop 1.13]. Thus, (the sum of) BGH and GHD is [also] less than two right-angles. But (straight-lines) being produced to infinity from (internal angles whose sum is) less than two right-angles meet together [Post. 5]. Thus, AB and CD, being produced to infinity, will meet together. But they do not meet, on account of them (initially) being assumed parallel (to one another) [Def. 1.23]. Thus, AGH is not unequal to GHD. Thus, (it is) equal. But, AGH is equal to EGB [Prop. 1.15]. And EGB is thus also equal to GHD. Let BGH be added to both. Thus, (the sum of) EGB and BGH is equal to (the sum of) BGH and GHD. But, (the sum of) EGB and BGH is equal to two right- 32 ETOIXEIfiN a'. ELEMENTS BOOK 1 [iepr] 8uoiv opdal? !aac oTcep e5ei 8eT?ai. angles [Prop. 1.13]. Thus, (the sum of) BGff and GHD is also equal to two right-angles. Thus, a straight-line falling across parallel straight- lines makes the alternate angles equal to one another, the external (angle) equal to the internal and opposite (an- gle), and the (sum of the) internal (angles) on the same side equal to two right-angles. (Which is) the very thing it was required to show. X'. Ai xfj auxfj eu'dsla TcapdXXr]Xoi xod dXXr|Xai<; eial TcapdXXr]- Xoi. Tiaxco sxaxspa x£>v AB, FA xfj EZ TcapdXXrjXoc;- Xey", on xod r) AB xrj TA eraxi TiapdXXrjXoc;. 'E^7iitix£xcl) yap slz auxag eu'deTa f] HK. Kod enel tic, TcapaXXf|Xou<; eu-Mac; xd<; AB, EZ su-dela e^KEnxcdxev f\ HK, tar] dpa f] utio AHK xfj utco H9Z. raxXiv, eitel eu; TcapaXXf|Xou<; euiDeia^ xd<; EZ, EA eu'dela e^KETixcdxev f\ HK, 'far) eaxlv f) utio H0Z xrj utco HKA. e8B)edr) 8e xal f] utio AHK xfj utco H9Z lor], xod f\ utco AHK dpa xfj utio HKA eaxiv I'ar)- xoa eiaiv evaXXd^. TiapdXXr]Xo<; dpa eaxlv f] AB xfj FA. [Ai dpa xfj auxfj eu-Ma TcapdXXr]Xoi xal dXXr|Xai<; rial jcapdXXr]Xoi-] OTcep e8ei 8ric;ai. Proposition 30 (Straight-lines) parallel to the same straight-line are also parallel to one another. Let each of the (straight-lines) AB and CD be parallel to EF. I say that AB is also parallel to CD. For let the straight-line GK fall across {AB, CD, and EF). And since the straight-line GK has fallen across the parallel straight-lines AB and EF, (angle) AGK (is) thus equal to GHF [Prop. 1.29]. Again, since the straight-line GK has fallen across the parallel straight-lines EF and CD, (angle) GHF is equal to GKD [Prop. 1.29]. But AGK was also shown (to be) equal to GHF. Thus, AGK is also equal to GKD. And they are alternate (angles). Thus, AB is parallel to CD [Prop. 1.27]. [Thus, (straight-lines) parallel to the same straight- line are also parallel to one another.] (Which is) the very thing it was required to show. Xa'. Aid xou 8oi9£vxo<; ar\{ieiou xfj Sorrier) eu-dsia uapdXXrjXov su-driav ypa^fjv dyayriv. 'Eaxw xo y.ev Scdev arjueTov xo A, i] Se Bcdriaa eu-dria f) Br- Sri Sf] Sid xou A ar^dou xfj BT eu'deia TiapdXXrjXov su-driav ypaji^fjv dyayriv. ElXfjcp-dw etc! xfj? BT xu)(6v arjuriov xo A, xal £TieC£U)fdw f] AA- xal auveaxdxco Tipog xfj AA eui9ria xal xw Tip6<; auxfj arjfieicp xw A xfj utio AAr y«v(a lor\ f\ utco AAE- xal Proposition 31 To draw a straight-line parallel to a given straight-line, through a given point. Let A be the given point, and BC the given straight- line. So it is required to draw a straight-line parallel to the straight-line BC, through the point A. Let the point D have been taken a random on BC, and let AD have been joined. And let (angle) DAE, equal to angle ADC, have been constructed on the straight-line 33 ETOIXEIfiN a'. ELEMENTS BOOK 1 sxpspXiqaiSw etc' eO'delac; xfj EA eu-Qsla f) AZ. Kal etcei etc; 8uo eu'deiac; xac; Br, EZ eu'dsTa e^nuTixouaa f] A A xac; evaXXdc; ytoviac; xac; utio EAA, AAr I'aac; dXXf]Xaic; TteTio(r]xev, TtapdXXr]Xoc; apa eaxlv f) EAZ xfj Br. Aid xoO 8o , devxoc; apa ar^eiou xoO A xfj Bo'deiar] cu-dela xfj Br TiapdXXrjXoc; eui9eTa ypa^turj ^ XTai ^ EAZ- oTtep eBei noifjaai. DA at the point A on it [Prop. 1.23]. And let the straight- line AF have been produced in a straight-line with EA. And since the straight-line AD, (in) falling across the two straight-lines BC and EF, has made the alternate angles EAD and ADC equal to one another, EAF is thus parallel to BC [Prop. 1.27]. Thus, the straight-line EAF has been drawn parallel to the given straight-line BC, through the given point A. (Which is) the very thing it was required to do. A(3'. Ilavxoc; xpiyovou [iiac; xQv TiXeupwv TipoaexpXrj'Marjc; f] exxoc; yiovia 8uol xau; evxoc; xal aTievavxiov lot) eaxlv, xod al evxoc; xoO xpiywvou xpsTc ycoviai 5ualv op'dalc; iaai eiaiv. A E B T A *Ectto xpiytovov xo ABr, xal TcpoaexpepXrja-dc) auxoO [lia TiXeupd f] Br era. xo A- Xeyw, oxi f] exxoc; ywvia f] utio ArA tar) eaxl 8ual xalc; evxoc; xal aTievavxiov xalc; utio TAB, ABr, xal ai evxoc; xoO xpiytovou xpeTc; ytovlai al utio ABr, BrA, TAB Bualv op^dic; iaai eiaiv. "Hy^co yap Bid xoO T arj^elou xfj AB eO'deia TiapdXXrjXoc; r I K. Kal etcsi TiapdXXrjXoc; eaxiv fj AB xfj TE, xal eic; auxdc; ejiTieKxcoxev f\ AT, al evaXXdi; ycovlai al utio BAr, ATE iaai dXXrjXaic; eiaiv. udXiv, enel TiapdXXrjXoc; eaxiv f) AB xfj TE, xal eic; auxdc; e^TieTixwxev eu-dela f) BA, f] exxoc; ytovla f) utio ErA iar) eaxl xfj evxoc; xal aTievavxiov xfj utio ABr. eSei/i}/) 5e xal f) utio ArE xfj utio BAr Tor)- oXrj dpa f\ utio ArA ycovia Iar) eaxl 8ual xdic; evxoc; xal aTievavxiov xalc; utco BAr, ABr. Proposition 32 In any triangle, (if) one of the sides (is) produced (then) the external angle is equal to the (sum of the) two internal and opposite (angles), and the (sum of the) three internal angles of the triangle is equal to two right-angles. A E BCD Let ABC be a triangle, and let one of its sides BC have been produced to D. I say that the external angle ACD is equal to the (sum of the) two internal and oppo- site angles CAB and ABC, and the (sum of the) three internal angles of the triangle — ABC, BCA, and CAB — is equal to two right-angles. For let CE have been drawn through point C parallel to the straight-line AB [Prop. 1.31]. And since AB is parallel to CE, and AC has fallen across them, the alternate angles BAC and ACE are equal to one another [Prop. 1.29]. Again, since AB is parallel to CE, and the straight-line BD has fallen across them, the external angle ECD is equal to the internal and opposite (angle) ABC [Prop. 1.29]. But ACE was also shown (to be) equal to BAC. Thus, the whole an- 34 ETOIXEIfiN a'. ELEMENTS BOOK 1 Koivf) TtpoaxsitrdtL) f\ Otto ATB- at apa utco ArA, ArB xpial xau; Otio ABr, BrA, TAB l'aai eiaiv. dXX' at bub ATA, ArB Sualv opflau; iaai eiaiv xal at utco ArB, TBA, TAB apa Sualv 6pi9aT<; laai eknv. Ilavxoc; apa xpiytovou ^iiac; xfiv xXeupfiv xpoaex- pXr^Bar^ f\ exxoc; ytovia Bual xaic; evxoc; xal dxevavxiov lar) eaxiv, xal at evxoc; xou xpiytovou xpeu; ywviai 8ualv opiate; laai eiaiv oxep eSei 8eTc;ai. gle ACD is equal to the (sum of the) two internal and opposite (angles) BAG and ABC. Let ACB have been added to both. Thus, (the sum of) ACD and ACB is equal to the (sum of the) three (angles) ABC, BCA, and CAB. But, (the sum of) ACL> and ACB is equal to two right-angles [Prop. 1.13]. Thus, (the sum of) ACB, CBA, and CAB is also equal to two right-angles. Thus, in any triangle, (if) one of the sides (is) pro- duced (then) the external angle is equal to the (sum of the) two internal and opposite (angles), and the (sum of the) three internal angles of the triangle is equal to two right-angles. (Which is) the very thing it was required to show. Ay. At xd<; laac; xe xal 7tapaXXf|Xouc; era. xd auxd ^epr] era- i^euyvuouaai euiJeTai xal auxal laai xe xal xapdXX/]Xoi eiaiv. B A A r "Eaxcoaav laai xe xal raxpdXX/jXoi at AB, TA, xal exi- i^euyvuxoaav aOxdc; era xd auxd uepr) eu-delai at AT, BA- Xeyo, oxi xal at Ar, BA laai xe xal xapdXXr)Xo[ eiaiv. 'Eue^eux'dw f] Br. xal eitet xapdXXr]X6c; eaxiv f] AB xfj TA, xal etc auxd<; eu-XCTixoxev f] Br, at evaXXdi; yoviai at uxo ABr, BrA laai dXXf|Xai<; eiaiv. xal exel Xor\ eaxlv f) AB xfj TA xoivf] 8e f] Br, 8uo Bf] at AB, Br 8uo xaic; Br, TA laai eiaiv xal ywvia f] bub AST ywvia xfj 0x6 BrA lary pdau; apa f] AT pdaei xfj BA eaxiv iar), xal xo ABr xpiyovov xcp BrA xpiywvw laov eaxiv, xal at Xoixal ywviai xaic; XoixaTc; ywviaic; laai eaovxai exaxepa exaxepa, Ocp' ac; at laai xXeupal O^oxetvouaiv iar) apa f] bub ATB yovia xfj 0x6 TBA. xal exel etc; 8uo eOiSeiac; xdc; AT, BA eO-dela ejixixxouaa f) Br xdc; evaXXdc; ywviac; laac; dXXfjXaic; xexo(/]xev, xapdXXrjXoc; apa eaxlv f] AT xfj BA. eSeix^r] 8e auxfj xal iar). At apa xdc; laac; xe xal 7iapaXXf]Xouc; era xd auxd \±£pr\ exii^euyvuouaai euiJeTai xal auxal laai xe xal xapdXX/]Xo( eiaiv Sxep e8ei 8eTc;ai. Proposition 33 Straight-lines joining equal and parallel (straight- lines) on the same sides are themselves also equal and parallel. B A D C Let AB and CD be equal and parallel (straight-lines), and let the straight-lines AC and BD join them on the same sides. I say that AC and BD are also equal and parallel. Let BC have been joined. And since AB is paral- lel to CD, and BC has fallen across them, the alter- nate angles ABC and BCD are equal to one another [Prop. 1.29]. And since AB is equal to CD, and BC is common, the two (straight-lines) AB, BC are equal to the two (straight-lines) DC, CBJ And the angle ABC is equal to the angle BCD. Thus, the base AC is equal to the base BD, and triangle ABC is equal to triangle DCB^, and the remaining angles will be equal to the corresponding remaining angles subtended by the equal sides [Prop. 1.4]. Thus, angle ACB is equal to CBD. Also, since the straight-line BC, (in) falling across the two straight-lines AC and BD, has made the alternate angles {ACB and CBD) equal to one another, AC is thus parallel to BD [Prop. 1.27]. And (AC) was also shown (to be) equal to (BD). Thus, straight-lines joining equal and parallel (straight- 35 ETOIXEIfiN a'. ELEMENTS BOOK 1 t The Greek text has "BC, CD", which is obviously a mistake. * The Greek text has "DCB", which is obviously a mistake. X8'. Tc5v 7iapaXXr]Xoypd[j.[jiiov ^topiwv al aTievavxiov TiXeupal xe xal ywvlai faai dXXr|Xaic; eiaiv, xal f\ Bidjiexpoc; auxa Si/a xe^vei. A B r a "Egxcl> 7tapaXX/]X6ypa[i^ov ^wpiov xo ArAB, Bid^iexpoc; 8e auxou f) Br- Xeyw, oxi xou ArAB TtapaXXrjXoypdjijjiou al dnevavxlov TiXeupal xe xal yovlai laai dXXr|Xai<; eiaiv, xal f] Br Bidjiexpoc; auxo 8()(a xe^tvei. Tkel yap napdXXr]X6<; eaxiv 7] AB xfj TA, xal sic, auxdg e^TCETixcoxEV euiDeTa rj Br, al evaXXai; ycovlai a ' 1 U7I ° ABr, BrA Taai aXX^Xaic; elalv. TtdXiv enel TiapdXXrjXog eaxiv f] Ar xfj BA, xal ei? auxdc; e^mcTixcoxev r] Br, al evaXXai; ycovlai al Otco ArB, TBA Taai dXXr|Xai<; elalv. 8uo 8r) xplycovd eaxi xd ABr, BrA xdc; 8uo ywviac; xd<; Otto ABr, BrA 8ual xdlc; bub BrA, TBA I'aag e)(ovxa exaxepav exaxepa xal ^tlav TtXeupdv [iia TtXeupa Tarjv x/]v Ttpoc; xdlc; Taai? ywviaic; xoivrjv auxov x/]v Br- xal xdc; Xomdc; apa TtXeupdc; xdlc; XoiTtdlc; Taac; ec;ei exaxepav exaxepa xal xrjv Xoircrjv ycovlav xfj XoiTtfj ywvia- I'ar) apa f) [lev AB TtXeupd xfj TA, f] 8e Ar xfj BA, xal exi Xat] eaxlv r] bub BAT yovla xfj Otio TAB. xal ensl Tar] eaxlv f] ^iev O716 ABr ycovia xfj bub BrA, rj 8e bub TBA xfj Oko ArB, oXr) apa f) Otco ABA oXr] xfj 0n:6 ArA eaxiv Xar]. eBelx'dr) 8e xal f] bub BAT xfj Otto TAB Tar]. Twv apa 7tapaXXr]Xoypd^a>v x w P^ wv a 'i aTievavxiov TiXeupal xe xal ywviai I'aai dXXrjXaic; elalv. Aeyw 8t], 6x1 xal f] Sid^texpoc; auxa Sixa xefivei. enel yap Tar] eaxlv f) AB xfj TA, xoivrj 8e f] Br, 860 Bf] al AB, Br 8ual xdlc; TA, Br Taai elalv exaxepa exaxepa- xal ywvla f) Otto ABr ytovla xfj Otto BrA la/], xal pdaic; apa f] AT xfj AB Tar], xal xo ABr [apa] xplycovov xw BrA xpiyovo Taov eaxlv. C H apa Br 8id(iexpo<; 8l)(a xejivei xo ABrA uapaX- Xf]kbypa.[i[Lov onep e8ei 8ele;ai. lines) on the same sides are themselves also equal and parallel. (Which is) the very thing it was required to show. Proposition 34 In parallelogrammic figures the opposite sides and angles are equal to one another, and a diagonal cuts them in half. A B C D Let ACDB be a parallelogrammic figure, and BC its diagonal. I say that for parallelogram ACDB, the oppo- site sides and angles are equal to one another, and the diagonal BC cuts it in half. For since AB is parallel to CD, and the straight-line BC has fallen across them, the alternate angles ABC and BCD are equal to one another [Prop. 1.29]. Again, since AC is parallel to BD, and BC has fallen across them, the alternate angles ACB and CBD are equal to one another [Prop. 1.29]. So ABC and BCD are two tri- angles having the two angles ABC and BCA equal to the two (angles) BCD and CBD, respectively, and one side equal to one side — the (one) by the equal angles and common to them, (namely) BC. Thus, they will also have the remaining sides equal to the corresponding re- maining (sides), and the remaining angle (equal) to the remaining angle [Prop. 1.26]. Thus, side AB is equal to CD, and AC to BD. Furthermore, angle BAG is equal to CDB. And since angle ABC is equal to BCD, and CBD to ACB, the whole (angle) ABD is thus equal to the whole (angle) ACD. And BAC was also shown (to be) equal to CDB. Thus, in parallelogrammic figures the opposite sides and angles are equal to one another. And, I also say that a diagonal cuts them in half. For since AB is equal to CD, and BC (is) common, the two (straight-lines) AB, BC are equal to the two (straight- lines) DC, CB\ respectively. And angle ABC is equal to angle BCD. Thus, the base AC (is) also equal to DB, 3G ETOIXEIfiN a'. ELEMENTS BOOK 1 and triangle ABC is equal to triangle BCD [Prop. 1.4]. Thus, the diagonal BC cuts the parallelogram ACDB^ in half. (Which is) the very thing it was required to show. t The Greek text has "CD, BC", which is obviously a mistake. * The Greek text has "ABCD", which is obviously a mistake. Xe'. Ta TtapaXXrjXoYpa^a xd era xfjc; aGxfjc; pdoewc; ovxa xal ev xalc; auxalc; TtapaXXfjXoic; Taa dXXf|Xoi<; eoxiv. A ' A E Z b r "Eaxw 7iapaXX/]X6Ypa[i^a xd ABTA, EBTZ era xfjc; auxfjc; pdoewc; xfjc; BT xal ev xau; auxau; TtapaXXrjXoic; xau; AZ, BT- Xsyw, oxi ioov laxl xo ABTA xG EBTZ TtapaXXr)- XoYpd^i^w. 'Etcei Ydp napaXXiqXoYpa^ov eoxi xo ABTA, Xar\ eoxlv f] AA xfj BT. 8id xd auxd 8f) xal f) EZ xfj BT eoxiv for] - Goxe xal #] AA xfj EZ eoxiv for) - xal xoivf] f] AE- oXrj dpa fj AE oXr) xfj AZ eoxiv for), eoxi 5e xal f] AB xfj AT for) - 8uo 8f) ai EA, AB Buo xau; ZA, AT I'aai eialv exaxepa exaxepa- xal yc^Via f) utto ZAT Y^via xfj bub EAB eoxiv I'or) T) exxoc; xfj evxoc;- pdou; dpa f] EB pdoei xfj ZT for) eoxiv, xal xo EAB xpiYiovov xG AZT xpiYGvip I'oov eoxai- xoivov dcprjprjadco xo AHE- Xoittov dpa xo ABHA xpaTte^iov Xomcp xG EHTZ xpanei^co eoxlv I'oov xoivov npoaxdadco xo HBT xpiYWvov 6Xov dpa xo ABTA 7iapaXXr)XoYpa|jiu.ov oXcp xG EBTZ 7rapaXXr)XoYpd^« I'oov eoxiv. Ta dpa TtapaXXr]X6Ypajj.[jia xd era xfjc; auxfjc; pdoewc; ovxa xal ev xalc; auxalc; 7tapaXXf|Xou; foa dXXfjXoic; eoxiv ouep e8ei SeT^ai. Proposition 35 Parallelograms which are on the same base and be- tween the same parallels are equal^ to one another. A D E F B C Let ABCD and EBCF be parallelograms on the same base BC, and between the same parallels AF and BC. I say that ABCD is equal to parallelogram EBCF. For since ABCD is a parallelogram, AD is equal to BC [Prop. 1.34]. So, for the same (reasons), EF is also equal to BC. So AD is also equal to EF. And DE is common. Thus, the whole (straight-line) AE is equal to the whole (straight-line) DF. And AB is also equal to DC. So the two (straight-lines) EA, AB are equal to the two (straight-lines) FD, DC, respectively. And angle FDC is equal to angle EAB, the external to the inter- nal [Prop. 1.29]. Thus, the base EB is equal to the base FC, and triangle EAB will be equal to triangle DFC [Prop. 1.4]. Let DGE have been taken away from both. Thus, the remaining trapezium ABCD is equal to the re- maining trapezium EGCF. Let triangle GBC have been added to both. Thus, the whole parallelogram ABCD is equal to the whole parallelogram EBCF. Thus, parallelograms which are on the same base and between the same parallels are equal to one another. (Which is) the very thing it was required to show. t Here, for the first time, "equal" means "equal in area", rather than "congruent". X<r'. Td napaXXrjXoYpa^a xd era focov pdoewv ovxa xal ev xalc; auxalc; napaXXfjXoic; foa dXXfjXoic; eoxiv. "Eot(x> 7iapaXX/]X6Ypa[i^a xd ABTA, EZH6 era focov pdoewv ovxa xGv BT, ZH xal ev xalc; auxalc; 7tapaXXf|Xou; xalc; A0, BH- Xcy«, oxi I'oov eaxi xo ABTA TtapaX- Proposition 36 Parallelograms which are on equal bases and between the same parallels are equal to one another. Let ABCD and EFGH be parallelograms which are on the equal bases BC and FG, and (are) between the same parallels AH and BG. I say that the parallelogram 37 ETOIXEIfiN a'. ELEMENTS BOOK 1 XrjXoYpaji^ov xcp EZH9. A A E BT Z H 'ETieCeu)fd«oav yap od BE, TO. xal ctccI iarj eaxlv f] Br xfj ZH, dXXd f] ZH xfj E0 eaxiv Tar), xal f) Br dpa xfj E9 eaxiv Xar\. eial 8s xal napdXXrjXoi. xal era^euyvuouaiv auxdc ai EB, OI 1, ai 8s xdc iaac xe xal TtapaXXfjXouc era xd auxd (iepr) era^euyvuouaai I'aai xe xal uapdXX/jXoi eiai [xal al EB, QT dpa Taai xe rim xal TiapdXXr)Xoi] . uapaX- Xr]X6ypa[i|jiov dpa eaxl xo EBr6. xai eaxiv I'aov ifi ABTA- pdaiv xe yap auxcp xfjv aOxfjv exei xf|v Br, xal ev xdic auxalc TtapaXXf]Xoic eaxlv auxc5 xau; Br, A0. 8la xd auxd 8rj xal xo EZH6 ifi auxcp xw EBr0 eaxiv I'aov waxe xal xo ABrA TtapaXXr]X6ypa[i(jiov ifi EZH6 eaxiv i'aov. Td dpa TtapaXXr]X6ypa[j.[jia xd era lawv pdaewv ovxa xal ev xalc auxalc; 7tapaXXf|Xoic laa dXXf|Xoic eaxiv orcep e8ei 8eu;ai. ABCD is equal to EFGH. AD EH B C F G For let BE and CH have been joined. And since BC is equal to FG, but FG is equal to EH [Prop. 1.34], BC is thus equal to EH. And they are also parallel, and EB and i?C join them. But (straight-lines) joining equal and par- allel (straight-lines) on the same sides are (themselves) equal and parallel [Prop. 1.33] [thus, EB and HC are also equal and parallel]. Thus, EBCH is a parallelogram [Prop. 1.34], and is equal to ABCD. For it has the same base, BC, as {ABCD), and is between the same paral- lels, BC and AH, as {ABCD) [Prop. 1.35]. So, for the same (reasons), EFGH is also equal to the same (par- allelogram) EBCH [Prop. 1.34]. So that the parallelo- gram ABCD is also equal to EFGH. Thus, parallelograms which are on equal bases and between the same parallels are equal to one another. (Which is) the very thing it was required to show. AC- Td xpiywva xd era xfjc auxfjc pdaewc ovxa xal ev xalc; auxalc napaXXfjXoic laa dXXfjXoic eaxiv. A A b r TCaxco xpiyova xd ABT, ABr era xfjc; auxfjc; pdaeoc xfjc Br xal ev xalc; auxalc TtapaXXf|Xoic xdic AA, BT- Xeyo, oxi laov eaxl xo ABT xpiywvov x£> ABT xpiyova). 'ExpepXfja'dM f] AA ecp' exdxepa xd [lipf] era xd E, Z, xal 8id \ism xou B xfj FA 7iapdXXr]Xoc fjx'dw f] BE, 8la 8e xou T xfj BA jiapdXXr]Xoc fj)edtL> f\ TZ. 7tapaXX/]X6ypa[i^.ov dpa eaxlv exdxepov xwv EBFA, ABTZ- xai riaiv iaa - era xe yap xfjc auxfjc pdaewc eloi xfjc Br xal ev xdic auxalc TtapaXXfjXoic xdic Br, EZ' xai eaxi xou ^.ev EBTA 7tapaXX/]Xoypd^ou fj^iiau xo ABT xpiywvov f] yap AB 8id[iexpoc auxo 8[)(a xe^vei- xou 8e ABTZ 7tapaXX/]Xoypd^ou fj[iiau xo ABr xpiycovov f) yap AT Sidjiexpoc auxo 8[)(a xejivei. [xd 8e Proposition 37 Triangles which are on the same base and between the same parallels are equal to one another. B C Let ABC and DBC be triangles on the same base BC, and between the same parallels AD and BC. I say that triangle ABC is equal to triangle DBC. Let AD have been produced in both directions to E and F, and let the (straight-line) BE have been drawn through B parallel to CA [Prop. 1.31], and let the (straight-line) CF have been drawn through C parallel to BD [Prop. 1.31]. Thus, EBCA and DBCF are both parallelograms, and are equal. For they are on the same base BC, and between the same parallels BC and EF [Prop. 1.35]. And the triangle ABC is half of the paral- lelogram EBCA. For the diagonal AB cuts the latter in 38 ETOLXEIftN a'. ELEMENTS BOOK 1 x£>v Tacov i]\iiay] Taa dXXr|Xoic; eaxiv] . Taov apa saxi to ABr xplywvov iw ABr xpiywvw. Td apa xplywva xd era xfj<; auxfj<; pdae«<; ovxa xal ev xau; auxau; 7iapaXXr|Xoi<; Taa dXXr]Xoi<; eaxiv oTtep eSei 8eTc;ai. t This is an additional common notion. XT)'. Td xpiyiova xd Era Tacov pdaecov ovxa xal ev xau; auxau; TtapaXXrjXou; Taa dXXrjXou; eaxiv. H A A BT E Z TEaxw xpiycova xd ABr, AEZ era Tacov pdaetov xwv Br, EZ xai ev xalg auxau; TtapaXXiqXou; xau; BZ, AA- Xeyto, oxi Taov eaxl xo ABr xpiyiovov xfi AEZ xpiyiovcp. 'ExpepXrja'dco yap f) AA ecp' exdxepa xd jiepr] era xd H, 0, xai Bid |iev xou B xf) TA TtapdXXr]Xoc; f])fdw f] BH, 8la 8s xou Z xfj AE 7iapdXX/]Xo<; f))fdw f] Z0. 7tapaXX/]X6ypa^ov apa eaxlv exdxepov xwv HBrA, AEZ6- xal Taov xo HBrA iS AEZ9- era xe yap Taov pdaewv eiai xov Br, EZ xal ev xau; auxau; TtapaXXr|Xou; xau; BZ, HO' xai eaxi xou [iev HBrA TtapaXXrjXoypd^iou r^uiau xo ABr xplywvov. f\ yap AB Bid^texpoc; auxo 8()(a xe^lvei' xou Be AEZ0 TtapaXXr]- Xoypd[i^tou 7]^iau xo ZEA xpiywvov rj yap AZ S(a^expo<; auxo 8()(a xe^tvei [xd Be xtov Tawv i]\±ioY) Taa aXXVjXou; eaxiv]. Taov apa eaxl xo ABr xplywvov xo AEZ xpiywvw. Td apa xpiywva xd era Tawv pdaetov ovxa xal ev xau; auxau; TtapaXXrjXou; Taa dXXf|Xoic; eaxiv oracp eSei 8eT<;ai. Td Taa xpiywva xd era xfj? auxrj<; pdaeioc; ovxa xal era xa auxd uepr) xal ev xau; auxau; 7iapaXXr|Xou; eaxiv. 'Eaxco Taa xpiywva xd ABr, ABr era xfj? auxfjg pdae«<; ovxa xal era xd auxd [iepi] xfj? Br- Xeyw, oxi xal ev xau; half [Prop. 1.34]. And the triangle DBC (is) half of the parallelogram DBCF. For the diagonal DC cuts the lat- ter in half [Prop. 1.34]. [And the halves of equal things are equal to one another.] t Thus, triangle ABC is equal to triangle DBC. Thus, triangles which are on the same base and between the same parallels are equal to one another. (Which is) the very thing it was required to show. Proposition 38 Triangles which are on equal bases and between the same parallels are equal to one another. G A D H B C E F Let ABC and DEF be triangles on the equal bases BC and EF, and between the same parallels BF and AD. I say that triangle ABC is equal to triangle DEF. For let AD have been produced in both directions to G and H, and let the (straight-line) BG have been drawn through B parallel to CA [Prop. 1.31], and let the (straight-line) FH have been drawn through F parallel to DE [Prop. 1.31]. Thus, GBCA and DEFH are each parallelograms. And GBCA is equal to DEFH. For they are on the equal bases BC and EF, and between the same parallels BF and GH [Prop. 1.36]. And triangle ABC is half of the parallelogram GBCA. For the diago- nal AB cuts the latter in half [Prop. 1.34]. And triangle FED (is) half of parallelogram DEFH. For the diagonal DF cuts the latter in half. [And the halves of equal things are equal to one another.] Thus, triangle ABC is equal to triangle DEF. Thus, triangles which are on equal bases and between the same parallels are equal to one another. (Which is) the very thing it was required to show. Proposition 39 Equal triangles which are on the same base, and on the same side, are also between the same parallels. Let ABC and DBC be equal triangles which are on the same base BC, and on the same side (of it). I say that 39 ETOIXEIfiN a'. ELEMENTS BOOK 1 auxaTc TtapaXXrjXou; eaxiv. Tke^eux'dto yap f) AA- Xeyw, 5xi TtapdXXr)X6c; eaxiv f] AA xrj BE. Ei yap [Lri, rj/iEko 8ia xoO A ar^eiou xfj BE eCWteia 7iapdXX/]Xo<; f] AE, xotl e7ie£eu)edio f) EE. I'oov apa eaxl xo ABr xpiytovov ifi EBr xpiytovor era xe yap T ^ auxrjc; pdaecoc; eaxiv auxco xfjc; Br xotl ev xdig auxdu; 7tapaXXf|Xoi<;. dXXd xo ABE xG ABr eaxiv laov xotl xo ABE apa xfi EBE laov eaxl xo uelCov ifi eXdaaovr ouep eaxiv d86vaxov oux apa TiapdXXr)X6<; eaxiv f] AE xfj BE. 6uo[m<; 8rj Sei^o^iev, oxi ou8' aXXr] tic; tiXtjv xfj? AA- f] AA apa xfj BE eaxi 7iapdXXr]Xo<;. Ta apa laa xpiywva xa era xfjc; auxfjc pdaeoc; ovxa xal era xa auxa ueprj xal ev xotu; auxau; 7iapaXXf|Xoi<; eaxiv oTtep e8ei 8eu;ai. Tiaxco laa xptyova xa ABE, EAE era latov pdaewv xwv Br, TE xal era xd aOxd uepr). Xeyto, oxi xal ev xdic auxalc TtapaXXr]Xoi<; eaxiv. 'Erae^eux'dco yap f] AA- Xeyw, oxi TtapdXXr]X6<; eaxiv f) AA xfj BE. Ei yap [Lr\, fj)fdw 8id xou A xrj BE TiapdXXr)Xo? f] AZ, xal s%eZ,si>x^ f] ZE. laov apa eaxl xo ABr xpiycovov xtp ZrE xpiytovor era xe yap lacov pdaetov eiai xtov Br, TE xal ev xalc auxau; TtapaXXr]Xou; xau; BE, AZ. dXXd xo ABE xpiycovov laov eaxl iu AEE [xpiycovcp] • xal xo AEE apa [xpiycovov] laov eaxl ifi ZEE xpiywvw xo ueTC^ov iw they are also between the same parallels. For let AD have been joined. I say that AD and BC are parallel. For, if not, let AE have been drawn through point A parallel to the straight-line BC [Prop. 1.31], and let EC have been joined. Thus, triangle ABC is equal to triangle EBC. For it is on the same base as it, BC, and between the same parallels [Prop. 1.37]. But ABC is equal to DBC. Thus, DBC is also equal to EBC, the greater to the lesser. The very thing is impossible. Thus, AE is not parallel to BC. Similarly, we can show that neither (is) any other (straight-line) than AD. Thus, AD is parallel to BC. Thus, equal triangles which are on the same base, and on the same side, are also between the same parallels. (Which is) the very thing it was required to show. Proposition 40^ Equal triangles which are on equal bases, and on the same side, are also between the same parallels. Let ABC and CDE be equal triangles on the equal bases BC and CE (respectively), and on the same side (of BE) . I say that they are also between the same par- allels. For let AD have been joined. I say that AD is parallel to BE. For if not, let AF have been drawn through A parallel to BE [Prop. 1.31], and let F E have been joined. Thus, triangle ABC is equal to triangle FCE. For they are on equal bases, BC and CE, and between the same paral- lels, BE and AF [Prop. 1.38]. But, triangle ABC is equal 40 ETOIXEIftN a'. ELEMENTS BOOK 1 eXdaaovi/ ouep eaxlv dBuvaxov oux apa TtapdXXr)Xo<; f) AZ xfj BE. 6(ioiCK 8r) 8eic;o^ev, oxi oi)B' dXXr] xu; TtXrjv xrjc; AA- f] AA apa xfj BE eaxi 7tapdXXr]Xo<;. Ta apa Taa xpiywva xa era iawv pdaewv ovxa xal ski xa auxa \±epr\ xal ev xau; auxdu; TtapaXXrjXou; eaxiv ouep eBei 8a^ai. to [triangle] DCE. Thus, [triangle] DCE is also equal to triangle FCE, the greater to the lesser. The very thing is impossible. Thus, AF is not parallel to BE. Similarly, we can show that neither (is) any other (straight-line) than AD. Thus, AD is parallel to BE. Thus, equal triangles which are on equal bases, and on the same side, are also between the same parallels. (Which is) the very thing it was required to show. t This whole proposition is regarded by Heiberg as a relatively early interpolation to the original text. \ia'. 'Edv TtapaXXiqXoypa^ov xpiywvtp pdaiv xe £)(/] xf|V auxrjv xal ev xal<; auxdu; TtapaXXrjXou; fj, BmXdaiov eaxi xo 7iapaXXr)X6ypa^ov xoO xpiytovou. A A E B r riapaXXr)X6Ypa^ov yap xo ABrA xpiywvcp xfi EBr pdaiv xe e^ex^ T, n v auxrjv xrjv Br xal ev xau; auxalc 7ta- paXXrjXou; eaxco xau; BT, AE- Xeyw, oxi SmXdaiov laxi xo ABrA TiapaXXrjXoypa^ov xou BEr xpiywvou. Tke^eux'dw yap f] Ar. Taov Sf| eaxi xo ABr xpiycovov x65 EBr xpiywvw- era xe yap xrj<; auxrj<; pdae«<; eaxiv auxCS xrj<; Br xal sv xau; auxdu; 7iapaXXr|Xou; xau; Br, AE. dXXd xo ABrA TtapaXXr)X6ypa^ov BiTtXdaiov "taxi xou ABr xpiyovou- f] yap Ar Sid^iexpo<; auxo Bixa xe^ver &axe xo ABrA 7iapaXX/]X6ypa[i^ov xal xou EBr xpiywvou laxl BiTtXdaiov. 'Edv apa KapaXXrjXoypa^iov xpiywvo pdaiv xe exTl T1 1 v auxr]v xal ev xau; auxdu; TtapaXXrjXou; fj, BiTtXdaiov eaxi xo 7iapaXX/]X6ypa[i^ov xou xpiywvou- ouep eBei 8eTc;ai. Tco BoiDevxi xpiywvw Taov TtapaXXr]X6ypa^ov auaxf]- aaaiSai sv xfj So'deiar) ywvia euiDuypd^^w. "Eaxco xo (lev BoiSev xpiycovov xo ABr, f) Be BoiDeToa ywvia eu-duypa^oi; rj A- Bel Br] xw ABT xpiywvw laov ua- paXXr)X6ypa^ov auaxrjaaa'dai ev xfj A ywvia eu-duypd^w. Proposition 41 If a parallelogram has the same base as a triangle, and is between the same parallels, then the parallelogram is double (the area) of the triangle. A D E B C For let parallelogram ABCD have the same base BC as triangle EBC, and let it be between the same parallels, BC and AE. I say that parallelogram ABCD is double (the area) of triangle BEC. For let AC have been joined. So triangle ABC is equal to triangle EBC. For it is on the same base, BC, as (EBC), and between the same parallels, BC and AE [Prop. 1.37]. But, parallelogram ABCD is double (the area) of triangle ABC. For the diagonal AC cuts the for- mer in half [Prop. 1.34]. So parallelogram ABCD is also double (the area) of triangle EBC. Thus, if a parallelogram has the same base as a trian- gle, and is between the same parallels, then the parallel- ogram is double (the area) of the triangle. (Which is) the very thing it was required to show. Proposition 42 To construct a parallelogram equal to a given triangle in a given rectilinear angle. Let ABC be the given triangle, and D the given recti- linear angle. So it is required to construct a parallelogram equal to triangle ABC in the rectilinear angle D. 41 ETOIXEIfiN a'. ELEMENTS BOOK 1 BET Tex^a'dco f] Br Bixa xaxd to E, xal era^eux'&co f] AE, xal auveaxdxo Ttpoc xfj Er euTJeia xal x£> npbc, auxrj ar^elw xo E xrj A ywvia Tar] f] utio TEZ, xal 81a (lev xou A xrj Er 7iapdXXr]Xo<; ^X'^ co ^ AH, 81a Be xoO T xfj EZ TiapdXXr]Xo<; f]xi}co f] TH- rcapaXXrjXoypa^ov dpa eaxl xo ZErH. xal end Xar\ eaxlv f) BE xrj Er, Taov eaxl xal xo ABE xptyovov tu AEr xpiywvw- era xe yap Tacov pdaecov elai x«v BE, Er xal ev xal? auxdig TiapaXXr|Xoi<; xau; Br, AH- BiuXdaiov dpa eaxl xo ABr xpiywvov xou AEr xpiyovou. eaxi Be xal xo ZErH 7iapaXXr)X6ypa[j.(jiov BiTtXdaiov xou AEr xpiywvou- pdaiv xe yap auxw xrjv auxrjv exei xal ev xau; auxau; eaxiv auxG TiapaXX^XoK;' I'aov dpa eaxl xo ZErH 7tapaXXr]X6ypa^ov iw ABr xpiywvw. xal e^ei xr]v uuo TEZ ycoviav Tarjv xrj Bo'deiar] xfj A. TO dpa Bo'devxi xpiywvw xw ABr Taov TiapaXXrjXoypa^t- \±ov auveaxaxai xo ZErH ev ywvia xrj Otto TEZ, f\ut; eaxlv Tar] xrj A- oitep eBei rcoirjaai. navxog TiapaXXrjXoypd^^iou xtov rcepl xr]v Bid^iexpov rca- paXXrjXoypd^wv xd KapaTiXr]p«^axa I'aa dXXr]Xoi<; eaxlv. 'Eaxw TtapaXXrjXoypa^ov xo ABrA, Bid^texpoc; Be auxou f) Ar, rcepl Be xrjv Ar TtapaXXr)X6ypa^(ia ^xev eaxo xd E9, ZH, xd Be Xeyojieva TiapaTiXr]p<^axa xd BK, KA- Xeyo, oxi Taov eaxl xo BK TcapaTtXrjpwpc xw KA Tiapa- TiXrjpwjiaxi. 'Excel yap rcapaXXr)X6ypa^6v eaxi xo ABrA, Bid^iexpoc; Be auxou f] AT, I'aov eaxl xo ABr xplyuvov x£> ArA xpiycovw. TidXiv, ercel TiapaXXrjXoypa^t^ov eaxi xo EG, Bid^iexpoc; Be auxou eaxiv f) AK, I'aov eaxl xo AEK xpiywvov tu A0K xpiywvw. 81a xd auxd Br] xal xo KZr xp(ya>vov xo KHr eaxiv I'aov. euel ouv xo jiev AEK xplyovov xo A6K xpiycovco eaxlv I'aov, xo Be KZr xt5 KHr, xo AEK xpiycovov ^xexa xou KHr Taov eaxl iu A0K xpiytovcp [icxd xou KZr- eaxi Be xal oXov xo ABr xplyovov 0X0 xc5 AAr laov Xoitiov dpa xo BK 7iapa7iXr]po^a XoikCS xw KA rcapa- B E C Let BC have been cut in half at E [Prop. 1.10], and let AE have been joined. And let (angle) CEF, equal to angle D, have been constructed at the point E on the straight-line EC [Prop. 1.23]. And let AG have been drawn through A parallel to EC [Prop. 1.31], and let CG have been drawn through C parallel to EF [Prop. 1.31]. Thus, FECG is a parallelogram. And since BE is equal to EC, triangle ABE is also equal to triangle AEC. For they are on the equal bases, BE and EC, and between the same parallels, BC and AG [Prop. 1.38]. Thus, tri- angle ABC is double (the area) of triangle AEC. And parallelogram FECG is also double (the area) of triangle AEC. For it has the same base as (AEC), and is between the same parallels as (AEC) [Prop. 1.41]. Thus, paral- lelogram FECG is equal to triangle ABC. (FECG) also has the angle CEF equal to the given (angle) D. Thus, parallelogram FECG, equal to the given trian- gle ABC, has been constructed in the angle CEF, which is equal to D. (Which is) the very thing it was required to do. Proposition 43 For any parallelogram, the complements of the paral- lelograms about the diagonal are equal to one another. Let ABCD be a parallelogram, and AC its diagonal. And let EH and FG be the parallelograms about AC, and BK and KD the so-called complements (about AC). I say that the complement BK is equal to the complement KD. For since ABCD is a parallelogram, and AC its diago- nal, triangle ABC is equal to triangle ACD [Prop. 1.34]. Again, since EH is a parallelogram, and AK is its diago- nal, triangle AEK is equal to triangle AHK [Prop. 1.34]. So, for the same (reasons), triangle KFC is also equal to (triangle) KGC. Therefore, since triangle AEK is equal to triangle AHK, and KFC to KGC, triangle AEK plus KGC is equal to triangle AHK plus KFC. And the whole triangle ABC is also equal to the whole (triangle) ADC. Thus, the remaining complement BK is equal to 42 ETOIXEIfiN a'. ELEMENTS BOOK 1 TiX/)p«|j.aT[ eaxiv igov. a e a b h r IlavToc; dpa TtapaXXrjXoYpd^ou /«p[ou xfiv itepl xf]v 8id^i£xpov 7iapaXX/]XoYpa[i^wv xa 7tapa7tXr)p«uaxa laa dXXr|- Xou; eaxiv OTtep s8ei 8eT^ai. the remaining complement KD. AH D B G C Thus, for any parallelogramic figure, the comple- ments of the parallelograms about the diagonal are equal to one another. (Which is) the very thing it was required to show. [i5'. ilocpd xrjv Bo'deiaocv eu'deiav xfi Sodevxi xpiYtovcp !aov 7ta- paXXrjXoYpa^ov TtapapaXeiv ev xfj SoOstor) ytovia eu'duYpd^i- [ICd. Z E K b h i — 7f A A "Eaxco f] ^jtev So-delaa eMeia f] AB, xo Be So'dev xpiywvov xo T, f] 8e bo'&eiaa ywvia eu'duYpa^tuoc; f] A- 8eT 5f] Ttapa xr]V Bo'deTaav sO'deiav xf]v AB xw Bcedevxi xpiYWvw xw T laov TtapaXXrjXoYpa^ov TtapapaXeTv ev Tar] xfj A Y^vta. Suveaxdxw x65 F xpiY^va) laov napaXX/jXoYpa^^ov xo BEZH ev Y«v[a xfj utio EBH, fj eaxiv Tar] xfj A- xa! xdcrdw oaxs en' euiSeiac; elvai xf)v BE xfj AB, xa! 5ir)X$co f] ZH era xo 0, xa! 81a xou A oiioxepa x£>v BH, EZ 7tapdXX/]Xo<; f^x'dw ^ A9, xa! ETC^euy^M f] 6B. xa! erce! etc; napaXXrjXouc; xdc; AG, EZ cu-deTa eveTteaev f) 9Z, ai dpa hub A0Z, 0ZE Ycoviai 8uaiv op'dau; eiaiv laai. a! dpa Otto BOH, HZE 860 bp-Q&v eXdaaovec; e£aiv ai 8s dno eXaaaovwv fj 860 opijfiv sic, dneipov expaXXojievai au^TUTixouaiv ai 6B, ZE Proposition 44 To apply a parallelogram equal to a given triangle to a given straight-line in a given rectilinear angle. F E K HA L Let AB be the given straight-line, C the given trian- gle, and D the given rectilinear angle. So it is required to apply a parallelogram equal to the given triangle C to the given straight-line AB in an angle equal to (angle) D. Let the parallelogram BEFG, equal to the triangle C, have been constructed in the angle EBG, which is equal to D [Prop. 1.42]. And let it have been placed so that BE is straight-on to AB) And let FG have been drawn through to H, and let AH have been drawn through A parallel to either of BG or EF [Prop. 1.31], and let HB have been joined. And since the straight-line HF falls across the parallels AH and EF, the (sum of the) an- gles AHF and HFE is thus equal to two right-angles 43 ETOIXEIfiN a'. ELEMENTS BOOK 1 dpa sx|3aXX6|jievai aujjuteaouvxai. expepXrjadwaav xal aufi- TUTixextoaav xaxd to K, xal 8id xou K arpeiou OTtoxepa x£5v EA, Z6 TiapdXXr]Xo<; f])(i9« f] KA, xal expepXr)aiL)Gjaav al 9A, HB erct xa A, M arj^eTa. TiapaXXr]X6ypa^ov apa eaxl to 6AKZ, Bidjiexpoc; 5e auxou f) 8K, Tiepi 8s xf)v 9K TiapaXXrjXoypajipia y.ev xa AH, ME, xa 8s Xeyo^teva Tiapa- 7iXr)pco[iaxa xa AB, BZ' iaov apa eaxl xo AB xfii BZ. dXXa xo BZ x£> r xpiycjvco eaxlv iaov xal xo AB apa xw T eaxiv 'iaov. xal etiel Xat] eaxlv f) utio HBE yiovia xfj utio ABM, dXXa f] utio HBE xfj A eaxiv Tar), xal f) utio ABM dpa xfj A ycovia eaxiv iar). napd xfjv SoiSeTaav dpa eui&elav xfjv AB x« So-devxi xpiycovco xco r i'aov TiapaXX/]X6ypa|jijj.ov TiapapepX/]xai xo AB ev yiovia xrj utio ABM, f] eaxiv Tar) xfj A- oTiep eSei Tioirjaai. t This can be achieved using Props. 1.3, 1.23, and 1.31. [IE. T£i Bo'devxi eu'duypd^w Iaov TiapaXXr]X6ypa[i^.ov auax- rjaaadai ev xfj Boifteiar) ywvia eu'duypd^iuw. "Eaxco xo ^tev So-dev eu'duypa^iuov xo ABrA, f] Be Bo'deTaa ywvia euiSuypa^uoc; f] E- 5eT §f] xw ABrA eO'du- ypd^tp iaov TiapaXXr)X6ypa^iuov auoxf]aaa'f)ai ev xfj Bo-deia?) yovia xfj E. 'ErceCsux'dw f] AB, xal auveaxdxo xw ABA xpiywvw iaov TtapaXXr]X6ypa^ov xo Z9 ev xrj utio 9KZ ywvia, f) eaxiv Tar] xfj E- xal TiapapepXf]adw Tiapa xf]v H9 euiMav iw ABr xpiycivo iaov TiapaXXr]X6ypa^ov xo HM ev xfj utio H9M ytovia, f\ eaxiv iar) xfj E. xal CTiei f) E ycovia exaxepa xGv utio 9KZ, H9M eaxiv Tar), xal f] utio 9KZ dpa xfj utio H9M eaxiv 'iar). xoivf] Tipoaxeia'dw f] utio K9H- ai dpa utio ZK9, K9H xdi<; utio K9H, H9M i'aai eiaiv. dXX' ai utio ZK9, K9H 8ualv opiate; Taai eiaiv xal ai utio K9H, H9M dpa Buo op'dau; Taai eiaiv. Tip6<; Sf] xivi eu'dela xfj H9 xal tu Tipoc auxfj ar^eio x<3 9 Buo euiSeTai ai K9, 9M [lt] era xa auxd [ispf] xei^ievai xa<; ecpe^fj<; y«via<; Buo opiJau; Taa<; Tioiouaiv tn euiJeiai; apa eaxlv f] K9 xfj 9M- xal cticI tic, TiapaXXf]Xou<; xd<; KM, ZH euiJeTa eveiieaev f) 9H, ai evaXXac; ycoviai ai utio M9H, 9HZ Taai dXXf|Xaic eiaiv. xoivf) Tipoaxeia'dcL) f] utio 9HA- ai dpa utio M9H, 9HA xau; utio 9HZ, 9 HA i'aai eiaiv. dXX' ai utio M9H, 9HA 8uo op'&dic, laai eiaiv xal ai utio 9HZ, 9HA apa Suo op^dit; laai eiaiv cti' euiDeiat; dpa eaxiv f) ZH xfj HA. xal cticI f) ZK xfj 9H Xar] xe xal TiapdXXr]X6<; eaxiv, dXXa xal f] 9H xfj MA, xal f) KZ apa xfj MA iar] xe xal TiapdXXr)X6<; eaxiv xal [Prop. 1.29]. Thus, (the sum of) BHG and GFE is less than two right-angles. And (straight-lines) produced to infinity from (internal angles whose sum is) less than two right-angles meet together [Post. 5]. Thus, being pro- duced, HB and FE will meet together. Let them have been produced, and let them meet together at K. And let KL have been drawn through point K parallel to either of EA or FH [Prop. 1.31]. And let HA and GB have been produced to points L and M (respectively) . Thus, HLKF is a parallelogram, and H K its diagonal. And AG and ME (are) parallelograms, and LB and BF the so-called complements, about HK. Thus, LB is equal to BF [Prop. 1.43]. But, BF is equal to triangle C. Thus, LB is also equal to C. Also, since angle GBE is equal to ABM [Prop. 1.15], but GBE is equal to D, ABM is thus also equal to angle D. Thus, the parallelogram LB, equal to the given trian- gle C, has been applied to the given straight-line AB in the angle ABM, which is equal to D. (Which is) the very thing it was required to do. Proposition 45 To construct a parallelogram equal to a given rectilin- ear figure in a given rectilinear angle. Let ABCD be the given rectilinear figure^ and E the given rectilinear angle. So it is required to construct a parallelogram equal to the rectilinear figure ABCD in the given angle E. Let DB have been joined, and let the parallelogram FH, equal to the triangle ABD, have been constructed in the angle HKF, which is equal to E [Prop. 1.42]. And let the parallelogram GM, equal to the triangle DBG, have been applied to the straight-line GH in the angle GHM, which is equal to E [Prop. 1.44]. And since angle E is equal to each of (angles) HKF and GHM, (an- gle) HKF is thus also equal to GHM. Let KHG have been added to both. Thus, (the sum of) FKH and KHG is equal to (the sum of) KHG and GHM. But, (the sum of) FKH and KHG is equal to two right-angles [Prop. 1.29]. Thus, (the sum of) KHG and GHM is also equal to two right-angles. So two straight-lines, KH and HM, not lying on the same side, make adjacent an- gles with some straight-line GH, at the point H on it, (whose sum is) equal to two right-angles. Thus, KH is straight-on to HM [Prop. 1.14]. And since the straight- line HG falls across the parallels KM and FG, the al- ternate angles MHG and HGF are equal to one another [Prop. 1.29]. Let HGL have been added to both. Thus, (the sum of) MHG and HGL is equal to (the sum of) 44 ETOIXEIfiN a'. ELEMENTS BOOK 1 era^suyvuoucuv auTa? eMelai ai KM, ZA- xai al KM, ZA apa I'aai te xal 7iapdXX/]Xo( elaiv 7tapaXX/]X6ypa^|iov dpa ecm to KZAM. xal end loov eoxl to ^iev ABA xp[y«vov to Z0 7iapaXX/)Xoypd|ji[j.cp, to 8s ABr tG HM, 6Xov dpa to ABrA eu^uypa^ov 6Xcp tc3 KZAM TTapaXXrjXoypdjUjiw screw iaov. A k e m Tco apa BoiDevTi Eu , duypd|i^.cp tw ABrA loov uapaX- Xr]X6ypa^ov auveoTaxai to KZAM ev yovia xfj Otio ZKM, f) eaxiv Tot) Tfj So'deior] xfj E- ouep e8si Tioifjaai. t The proof is only given for a four-sided figure. However, the extension Ako Tfj? 8o , de(or]<; su-delac. xerxpdywvov dvaypdtjjai. 'Eoxo #] Bo-fMaa euTMa f] AB- 8eT 5f| duo xrjc AB eO'deiac; xexpdycovov dvaypdtjiai. "Hy^co xfj AB euiiteia diio toO Tipoc. auxfj ar)[ie(ou xou A itpoc. opM? f\ AT, xal xeicrdco Tfj AB lot] f\ AA- xal 8ia ^iev tou A ar)[ie(ou xfj AB 7iapdXXr]Xoc. fix^to f] AE, 8ia 8e toO B ar)[i£Lou Tfj AA 7tapdXXr]Xo<; rjx$co f] BE. napaX- X/]X6ypa[i^ov apa eaxl to AAEB- tar] apa soxlv f] ^isv AB Tfj AE, f] 8e AA Tfj BE. dXXa f] AB Tfj AA eaxiv larj- ai xeaaape? dpa ai BA, AA, AE, EB '{aai dXXf]Xaic sicnv iaoTtXsupov apa ecru to AAEB TtapaXXr)X6ypa[i(jiov. Xeyco 8r], otl xal opiJoyoviov. enel yap el? uapaXX^Xou? xd? AB, AE su'dela eveueaev f\ AA, al dpa (mo BAA, AAE ycovtai 8uo op-ddl? '(aai eialv. dpfty) 8e f] Otio BAA- op'df] dpa xal HGF and HGL. But, (the sum of) MHG and ffGL is equal to two right-angles [Prop. 1.29]. Thus, (the sum of) HGF and i/GL is also equal to two right-angles. Thus, FG is straight-on to GL [Prop. 1.14]. And since FK is equal and parallel to EG [Prop. 1.34], but also HG to ML [Prop. 1.34], KF is thus also equal and parallel to ML [Prop. 1.30]. And the straight-lines KM and FL join them. Thus, KM and FL are equal and parallel as well [Prop. 1.33]. Thus, KFLM is a parallelogram. And since triangle ABD is equal to parallelogram FH, and DBC to GM, the whole rectilinear figure ABCD is thus equal to the whole parallelogram KFLM. D K H M Thus, the parallelogram KFLM, equal to the given rectilinear figure ABCD, has been constructed in the an- gle FKM, which is equal to the given (angle) E. (Which is) the very thing it was required to do. to many-sided figures is trivial. Proposition 46 To describe a square on a given straight-line. Let AB be the given straight-line. So it is required to describe a square on the straight-line AB. Let AC have been drawn at right-angles to the straight-line AB from the point A on it [Prop. 1.11], and let AD have been made equal to AB [Prop. 1.3]. And let DE have been drawn through point D parallel to AB [Prop. 1.31], and let BE have been drawn through point B parallel to AD [Prop. 1.31]. Thus, ADEB is a parallelogram. Therefore, AB is equal to DE, and AD to BE [Prop. 1.34]. But, AB is equal to AD. Thus, the four (sides) BA, AD, DE, and EB are equal to one another. Thus, the parallelogram ADEB is equilateral. So I say that (it is) also right-angled. For since the straight-line 45 ETOIXEIfiN a'. ELEMENTS BOOK 1 f) Otio AAE. twv 8e 7iapaXX/]Xoypd|jijj.(x>v )(«p[wv ai dme- vavxiov TtXeupai xe xal ywviai i'aai dXXfjXaic; eiaiv op-df) dpa xai exaxepa xc5v dnevavxiov xGv utio ABE, BEA ycovicov op^oycoviov apa eaxl to AAEB. eSeix'dr] 8e xal iaoftXeupov. r A E A B Texpdywvov apa eaxiv xa[ eaxiv arco xfjc; AB eu-deiac; dvayeypa^ievov onep eSei Troifjaai. 'Ev xolc; op^oyMvioic; xpiycovoig xo duo xfjc; xfjv op^rjv ycoviav un;oxeivouar]<; TtXeupac; xexpdywvov laov eaxl xou; duo iwv xf]V 6p$r]v ycoviav Ttepiexouawv TtXeupwv xe- xpaywvou;. "Eaxto xpiywvov op-doycoviov xo ABr op^/jv exov xrjv uno BAr ycoviav Xeyo, oxi xo duo xfjc; Br xexpdycovov Taov eaxl xolc dfto xcov BA, Ar xexpaywvoic;. Avayeypdcp'dw yap and y.ev xfjc; Br xexpdyiovov xo BAEr, duo 8e xwv BA, Ar xd HB, 9r, xai Bid xoO A oTtoxepa xwv BA, TE 7tapdXXr]Xoc; fix^" f) AA- xal ETieCeux'dwaav ai AA, ZT. xal ind 6pi5f| eaxiv exaxepa iwv utio BAr, BAH ywviwv, Ttpoc; 8f] xivi cu-Ma xfj BA xal x£S Ttpoc; auxfj arjueiM ifi A 860 eui9eTai ai Ar, AH ^tf) excl xd auxd \iepf) xei^ievai xac; ecpec;fjc; ycoviac; 8ualv opiJaTc; Taac; TtoioOaiv z% eO'deiac; apa eaxiv f] FA xfj AH. 81a xd auxd 8f) xal f) BA xfj A0 eaxiv tn eO'deiac;. xal enel tar] eaxlv f] O716 ABr ywvia xfj uko ZBA- opiSf] yap exaxepa- xotvf) npoaxeia'dw f] tmo ABr- oXrj apa f\ hub ABA oXr] xfj O716 ZBr eaxiv lay), xal ckcI iar) eaxlv f) ^xev AB xfj Br, f) 8e ZB xfj BA, 860 8f] ai AB, BA 860 xdic; ZB, Br laai eiaiv exaxepa exaxepa- xai ywvia f] 6716 ABA yovia xfj bub ZBr Tar) - pdaic; apa f] AA pdaei xfj Zr [eaxiv] Iar], xal xo ABA AD falls across the parallels AB and DE, the (sum of the) angles BAD and ADE is equal to two right-angles [Prop. 1.29]. But BAD (is a) right-angle. Thus, ADE (is) also a right-angle. And for parallelogrammic figures, the opposite sides and angles are equal to one another [Prop. 1.34]. Thus, each of the opposite angles ABE and BED (are) also right-angles. Thus, ADEB is right- angled. And it was also shown (to be) equilateral. c D E A B Thus, {ADEB) is a square [Def. 1.22]. And it is de- scribed on the straight-line AB. (Which is) the very thing it was required to do. Proposition 47 In right-angled triangles, the square on the side sub- tending the right-angle is equal to the (sum of the) squares on the sides containing the right-angle. Let ABC be a right-angled triangle having the angle BACa right-angle. I say that the square on BC is equal to the (sum of the) squares on BA and AC. For let the square BDEC have been described on BC, and (the squares) GB and HC on AB and AC (respectively) [Prop. 1.46]. And let AL have been drawn through point A parallel to either of BD or CE [Prop. 1.31]. And let AD and FC have been joined. And since angles BAG and BAG are each right-angles, then two straight-lines AC and AG, not lying on the same side, make the adjacent angles with some straight-line BA, at the point A on it, (whose sum is) equal to two right-angles. Thus, CA is straight-on to AG [Prop. 1.14]. So, for the same (reasons), BA is also straight-on to AH. And since angle DBC is equal to FBA, for (they are) both right-angles, let ABC have been added to both. Thus, the whole (angle) DBA is equal to the whole (an- gle) FBC. And since DB is equal to BC, and FB to BA, the two (straight-lines) DB, BA are equal to the 4G ETOIXEIfiN a'. ELEMENTS BOOK 1 xpiywvov xG ZBr xpiywvw saxlv iaov xa( [eaxi] xou ^.ev ABA xpiyovou 8iitXdaiov xo BA TiapaXXr)X6ypa^i|iov pdaiv xe yap xfjv auxfjv s^ouai xf]v BA xal ev xau; auxau; elai 7tapaXXf]Xoi<; xau; BA, AA - xou Se ZBr xpiywvou 8i7tXdaiov xo HB xexpdywvov pdaiv xe yap ndXiv xf]v auxrjv e)(ouai xf]v ZB xal ev xau; auxau; eiai TtapaXXf|Xou; xau; ZB, Hr. [xa 8e xwv iaov 8iTiXdaia laa dXXf|Xou; eaxiv] iaov apa eaxi xal xo BA KapaXXrjXoypa^^iov xw HB xexpaywvw. 6[ioiok 8f] eTuCeuyvu^evwv xwv AE, BK Beix^rpexai xal xo TA 7tapaXX/]X6ypa^ov I'aov iw 0T xexpaywvy oXov apa xo BAEr xexpdywvov 8uol xou; HB, 8r xexpaywvou; iaov eaxiv. xal eaxi xo ^iev BAEr xexpdycovov drco xfjc Br dva- ypacpev, xa 8e HB, OT and xcov BA, AI\ xo apa anb xfjc; Br TtXeupac; xexpdyovov iaov eaxi xolc duo xfiv BA, Ar uXeupov xexpayovou;. A A E 'Ev apa xou; op'doyovioic; xpiywvoic; xo dno xfjc; xfjv op'dfjv ywviav UTtoxeivouarjc; TtXeupdc; xexpdyiovov i'aov eaxi xoTc; duo xGv xfjv 6pi3f)v [ywviav] rcepiexouacov TtXeupcov xe- xpaywvoic oTtep eSei Sel^ai. t The Greek text has "FB, BC", which is obviously a mistake, t This is an additional common notion. two (straight-lines) CB, BF,^ respectively. And angle DBA (is) equal to angle FBC. Thus, the base AD [is] equal to the base FC, and the triangle ABD is equal to the triangle FBC [Prop. 1.4]. And parallelogram BL [is] double (the area) of triangle ABD. For they have the same base, BD, and are between the same parallels, BD and AL [Prop. 1.41]. And square GB is double (the area) of triangle FBC. For again they have the same base, FB, and are between the same parallels, FB and GC [Prop. 1.41]. [And the doubles of equal things are equal to one another.]- 1 - Thus, the parallelogram BL is also equal to the square GB. So, similarly, AE and BK being joined, the parallelogram CL can be shown (to be) equal to the square HC. Thus, the whole square BDEC is equal to the (sum of the) two squares GB and HC. And the square BDEC is described on BC, and the (squares) GB and HC on BA and AC (respectively) . Thus, the square on the side BC is equal to the (sum of the) squares on the sides BA and AC. H D L E Thus, in right-angled triangles, the square on the side subtending the right-angle is equal to the (sum of the) squares on the sides surrounding the right- [angle]. (Which is) the very thing it was required to show. 47 ETOIXEIfiN a'. ELEMENTS BOOK 1 'Edv xpiycovou to and [iiag xfiv TiXeupfiv xexpdycovov laov fj xolc; duo xfiv Xoittcov xou xpiycovou 860 TtXeupcov xexpaycovou;, f) Tt£pi£)(o|jievr] ycovia u^o Xomcov xou xpiycovou 860 TtXeupcov opfir] saxiv. A A B Tpiycovou yap xou ABr xo dno [iiac, xfjg Br TtXeupac; xexpdycovov l'aov saxco xou; dno xcov BA, Ar TtXeupcov xe- xpaycovou;- Xeyco, 6x1 op'dr] eaxiv f) utio BAr ycovia. "H)edco yap diio xou A ar\\ieiou xrj Ar eu'deia Tipo<; 6pi5d<; f] A A xal xeiaiJco xfj BA Tar] f) AA, xal eTieCeuyiJco f] Ar. End Tar] eaxlv f) A A xfj AB, laov eaxl xal xo duo xrjc AA xexpdycovov xo duo xfj? AB xexpaycovcp. xoivov Tipo- axeiai9co xo and xrjc; Ar xexpdycovov xd dpa duo xcov AA, Ar xexpdycova Taa eaxl xou; arco xcov BA, Ar xexpaycovou;. dXXd xou; [iev and xcov AA, Ar Taov eaxl xo ano xrjg AE 6pTL>r) yap eaxiv f) utio AAr ycovia- xou; 8s duo xcov BA, Ar laov eaxl xo duo xfjc; Br- UTtoxeixai yap- xo dpa arco xrjc; Ar xexpdycovov laov sax! xco dno xrjc; Br xexpaycovcp- coaxe xal TtXeupa f) Ar xrj Br eaxiv Iar]- xal end Tar) eaxlv f) AA xfj AB, xoivr) 8s f] AT, Suo 8/) ai AA, Ar 660 xdu; BA, Ar Taai eiaiv xal pdaic; f) Ar pdaei xfj Br lot)- ycovia dpa f) utio AAr ycovia xrj utio BAr [eaxiv] Iar). 6pi5r] 8e f] utio AAr- op-Qr) dpa xal f] utio BAr. 'Edv dpa xpiycovou xo duo [ii&c, xcov TtXeupcov xexpdycovov laov fj xolc; duo xcov Xoiticov xou xpiycovou 860 TtXeupcov xexpaycovou;, f) Ttepie/o|jievr] ycovia utio xcov Xoiticov xou xpiycovou 860 TtXeupcov op-dr] eaxiv oTtep eSei SeT^ai. Proposition 48 If the square on one of the sides of a triangle is equal to the (sum of the) squares on the two remaining sides of the triangle then the angle contained by the two remain- ing sides of the triangle is a right-angle. C DAB For let the square on one of the sides, BC, of triangle ABC be equal to the (sum of the) squares on the sides BA and AC. I say that angle BAC is a right-angle. For let AD have been drawn from point A at right- angles to the straight-line AC [Prop. 1.11], and let AD have been made equal to BA [Prop. 1.3], and let DC have been joined. Since DA is equal to AB, the square on DA is thus also equal to the square on AB) Let the square on AC have been added to both. Thus, the (sum of the) squares on DA and AC is equal to the (sum of the) squares on BA and AC. But, the (square) on DC is equal to the (sum of the squares) on DA and AC. For an- gle DAC is a right-angle [Prop. 1.47]. But, the (square) on BC is equal to (sum of the squares) on BA and AC. For (that) was assumed. Thus, the square on DC is equal to the square on BC. So side DC is also equal to (side) BC. And since DA is equal to AB, and AC (is) com- mon, the two (straight-lines) DA, AC are equal to the two (straight-lines) BA, AC. And the base DC is equal to the base BC. Thus, angle DAC [is] equal to angle BAC [Prop. 1.8]. But DAC is a right-angle. Thus, BAC is also a right-angle. Thus, if the square on one of the sides of a triangle is equal to the (sum of the) squares on the remaining two sides of the triangle then the angle contained by the re- maining two sides of the triangle is a right-angle. (Which is) the very thing it was required to show. t Here, use is made of the additional common notion that the squares of equal things are themselves equal. Later on, the inverse notion is used. 48 ELEMENTS BOOK 2 Fundamentals of Geometric Algebra STOIXEIQN p\ ELEMENTS BOOK 2 "Opoi. a'. Ilav TTapaXXr)X6ypa[j.(jiov opi&oywviov Tiepiexea-dai Xeyexai utio Buo xfiv xf]v opiS/jv yiovlav TiepiexouaCiv eMeifiv. P'. IlavToc; 8s TiapaXX/]Xoypd[ji^ou x^plou iSv rcepl xf)v 8idpiexpov auxoO TiapaXXrjXoypd^ojv ev ottoiovouv auv xoT<; 5ua! TiapaTiXr)p«|jiaai yvw^wv xaXeladio. Definitions 1. Any rectangular parallelogram is said to be con- tained by the two straight-lines containing the right- angle. 2. And in any parallelogrammic figure, let any one whatsoever of the parallelograms about its diagonal, (taken) with its two complements, be called a gnomon. 'Edv Sai 8uo eOdelai, Tji.rj'dfj 8e f] exepa auxwv etc; oaa- 8r)7ioxoOv x^ir][iaxa, xo Tiepiexojievov op^oycoviov utio xwv 8uo eO'dei&v laov eaxl xou; utio xe xrjc; dx[if|xou xod exdaxou x£>v x^ir^dxtov Tiepiexo[ievoi<; op-doycovioic;. A- B K A H Z TEaxoaav 8uo einMai ai A, Br, xal xex^o'dw f) Br, cbc; exuxev, xaxd xd A, E ar)^eTa- Xeyw, oxi xo utio twv A, Br Tiepiexo|ievov opiSoytoviov laov eaxl xco xe utio xc5v A, BA Tiepiexo^ievtp op^oywviw xal xw utio xwv A, AE xal exi xfi utio x«v A, EE "H)cdco yap and xou B xfj Br Tipog opiJdc; f) BZ, xal xeladio xfj A for) f] BH, xal Sid [Lev xou H xfj Br TiapdXXr]Xo<; rjX'dw f] H0, 8id 8e xwv A, E, T xfj BH TiapdXX/]Xoi fjx'dwaav aiAK, EA, TO. 'laov 8f| eaxi xo B9 xoI<; BK, AA, E0. xal eaxi xo ^iev B6 xo utio xwv A, Br- Tiepiexexai jiev yap utio xwv HB, Br, I'ar) 8e f\ BH xfj A- xo 8e BK xo utio xwv A, BA- Tiepiexexai ^xev yap utio xwv HB, BA, iar) 8e f] BH xfj A. xo 8e AA xo utio xwv A, AE- i'ar) yap f) AK, xouxeaxiv f] BH, xfj A. xal exi ojioiclx; xo E0 xo utio xwv A, EE xo apa utio x£Sv A, Br laov eaxl xw xe utio A, BA xal xo utio A, AE xal exi tu utio A, EE 'Edv apa Sai 8uo eu'delai, xpydfj Be f] exepa auxGv tic, 6aa8/]Tioxouv x[if]^axa, xo Tiepiexopievov op'doyoviov utio xwv Buo eu-deiwv I'aov eaxl xou; utio xe xfj<; dx^xou xal exdaxou xGv x[i/]^dxwv Tiepiexo^evou; op-doyMvlou;- oTiep Proposition 1+ If there are two straight-lines, and one of them is cut into any number of pieces whatsoever, then the rectangle contained by the two straight-lines is equal to the (sum of the) rectangles contained by the uncut (straight-line), and every one of the pieces (of the cut straight-line) . A B D K H Let A and BC be the two straight-lines, and let BC be cut, at random, at points D and E. I say that the rect- angle contained by A and BC is equal to the rectangle(s) contained by A and BD, by A and DE, and, finally, by A and EC. For let BF have been drawn from point B, at right- angles to BC [Prop. 1.11], and let BG be made equal to A [Prop. 1.3], and let GH have been drawn through (point) G, parallel to BC [Prop. 1.31], and let DK, EL, and CH have been drawn through (points) D, E, and C (respectively), parallel to BG [Prop. 1.31]. So the (rectangle) BH is equal to the (rectangles) BK, DL, and EH. And BH is the (rectangle contained) by A and BC. For it is contained by GB and BC, and BG (is) equal to A. And BK (is) the (rectangle contained) by A and BD. For it is contained by GB and BD, and BG (is) equal to A. And DL (is) the (rectangle contained) by A and DE. For DK, that is to say BG [Prop. 1.34], (is) equal to A. Similarly, EH (is) also the (rectangle con- tained) by A and EC. Thus, the (rectangle contained) by A and BC is equal to the (rectangles contained) by A 50 STOIXEIQN p\ ELEMENTS BOOK 2 e8ei 8eT<;ou. and BD, by A and DE, and, finally, by A and EC. Thus, if there are two straight-lines, and one of them is cut into any number of pieces whatsoever, then the rectangle contained by the two straight-lines is equal to the (sum of the) rectangles contained by the uncut (straight-line), and every one of the pieces (of the cut straight-line) . (Which is) the very thing it was required to show. t This proposition is a geometric version of the algebraic identity: a(b + c + c2 H ) = ab + ac + ad + ■ P'- 'Edv eO'dsTa ypa^iuf] xur^fj, cdc, exu)(ev, to Otto xfjc; 6kr\<z xal exaxepou xGv xur^dxcov Ttepie)(6pievov opiJoycoviov iaov eaxl xo arco xfjc; oArjc; xexpayovcp. a r B Proposition 2+ If a straight-line is cut at random then the (sum of the) rectangle (s) contained by the whole (straight-line), and each of the pieces (of the straight-line), is equal to the square on the whole. AC B A Z E Eu'deTa yap f) AB xex[ifjad<j, ok exu)(ev, xaxa xo T o/)[ieTov Xeyo, oxi xo utco xwv AB, Br Tcepie)(6^evov op^oyMviov jiexd xou utco BA, Ar Tcepie)(o|jievou op-do- yoviou laov eaxl xw duo xfjc; AB xexpaytovio. Avayeypdcp'do yap diro xfjc; AB xexpdytovov xo AAEB, xal fix&oi Sid xou T oTcoxepa xwv AA, BE TiapdXXrjXoc; f) rz. 'laov Sfj eaxl xo AE xou; AZ, TE. xa( eaxi xo ^.ev AE xo dico xfjc; AB xexpdywvov, xo 8s AZ xo Otco xGv BA, Ar Tiepie)(o^i£vov opTSoytoviov Tcepie)(exai ydp otco xwv A A, Ar, lar\ 8e f) A A xfj AB- xo 8e TE xo Otco xfiv AB, Br- iar) yap f) BE xfj AB. xo dpa utco xfiv BA, Ar jiexa xou utco xtov AB, Br laov eaxl xw aTco xfjc; AB xexpayova). 'Eav dpa eu'de'ia ypa^f) x^trydfj, wc; exu)(ev, xo utco xfjc; oX/]c; xal exaxepou xwv x^irj^dxwv Tcepie)(6^evov op'doywvi.ov laov eaxl x£5 aTco xfjc; oXrjc; xexpaywvw- OTcep e8ei 8eu;ai. D F E For let the straight-line AB have been cut, at random, at point C. I say that the rectangle contained by AB and BC, plus the rectangle contained by BA and AC, is equal to the square on AB. For let the square ADEB have been described on AB [Prop. 1.46], and let CF have been drawn through C, parallel to either of AD or BE [Prop. 1.31]. So the (square) AE is equal to the (rectangles) AF and CE. And AE is the square on AB. And AF (is) the rectangle contained by the (straight-lines) BA and AC. For it is contained by DA and AC, and AD (is) equal to AB. And CE (is) the (rectangle contained) by AB and BC. For BE (is) equal to AB. Thus, the (rectangle con- tained) by BA and AC, plus the (rectangle contained) by AB and BC, is equal to the square on AB. Thus, if a straight-line is cut at random then the (sum of the) rectangle (s) contained by the whole (straight- line), and each of the pieces (of the straight-line), is equal to the square on the whole. (Which is) the very thing it was required to show. 51 STOIXEIQN p\ ELEMENTS BOOK 2 t This proposition is a geometric version of the algebraic identity: a b + Y • 'Edv eO'dsTa ypa[iur] T^y^df), cbc, exuxev, to utco xrjc oXrjc xal evoc, xGv xjir^dxtov Tcepiexo^ievov opiDoyoviov iaov sail to xe utco tov x^ir)^dxcov Tcepiexo^iivw opTJoyMvltp xal iS aTco toD Tcpoeipr)^evou xfjirinaxoc, xexpaytovw. a r b Z A E EO'deTa yap f] AB xex^yja'dto, <l>c exuxev, xaxd to E Xeyco, oxi to utco twv AB, Br Tcepiexo^ievov opiSoycoviov iaov eaxl xfi xe Otto twv Ar, TB Tcepiexo^ievw opiJoyovia) ^.exd tou dico xrjc, Br xexpaywvou. AvayeypdcpTJco Y^P aTI ° -TB xexpdywvov to TAEB, xal BirjX'dw ttj EA em to Z, xal Sid tou A oTcoxepa xcov FA, BE TcapdXXr)Xoc, f)x&M f\ AZ. iaov Sr] eaxi to AE tou; AA, TE' xal 6cm to [lzv AE to utco xtov AB, Br Tcepiex6[ievov opiDoycoviov Tcepiexexai [Lev yap utco iuv AB, BE, iar\ Se f] BE xfj Br- to 8s AA to utco tuv Ar, TB- lot] yap f\ AT xfj TB' to 8e AB to duo xfjc TB xexpaycovov xo apa utco xwv AB, Br Tcepiexo^tevov opiDoycoviov laov eaxl ifi utco icov Ar, TB Tcepiexo^tevw op-doycovlo jiexd tou aTco xfj? Br xexpaytovou. 'Eav apa eu-dela ypajji^tr] T[a.rj'df), cbc; exuxev, xo utco xrjc, oXr]C. xal evoc. xfiv x^irj^dxwv Tcepiexojievov opiJoycoviov laov eaxl xO xe utco xtov xjir^dxwv Tcepiexo^ievtL> op-doytovlo xal x<3 dTco xou Tcpo£ipr)[ievou x^rpaxoc, xexpaycovcp- oTcep eSei 8eT<;ai. t This proposition is a geometric version of the algebraic identity: (a + 5'. 'Eav eu'deTa yp°Wn "■W^ exuxev, xo duo xfjc; oXr]C_ xexpdywvov laov eaxl xolc xe aTco xwv x^r)[idx«v xe- xpaycovoic. xal xQ 81? utco xwv x^tr]udxov Tcepiexo^tevw op'do- c = a 2 if a = b + c. Proposition 3 1 " If a straight-line is cut at random then the rectangle contained by the whole (straight-line), and one of the pieces (of the straight-line), is equal to the rectangle con- tained by (both of) the pieces, and the square on the aforementioned piece. AC B F D E For let the straight-line AB have been cut, at random, at (point) C. I say that the rectangle contained by AB and BC is equal to the rectangle contained by AC and CB, plus the square on BC. For let the square CDEB have been described on CB [Prop. 1.46], and let ED have been drawn through to F, and let AF have been drawn through A, parallel to either of CD or BE [Prop. 1.31]. So the (rectangle) AE is equal to the (rectangle) AD and the (square) CE. And AE is the rectangle contained by AB and BC. For it is contained by AB and BE, and BE (is) equal to BC. And AD (is) the (rectangle contained) by AC and CB. For DC (is) equal to CB. And DB (is) the square on CB. Thus, the rectangle contained by AB and BC is equal to the rectangle contained by AC and CB, plus the square on BC. Thus, if a straight-line is cut at random then the rect- angle contained by the whole (straight-line), and one of the pieces (of the straight-line), is equal to the rectangle contained by (both of) the pieces, and the square on the aforementioned piece. (Which is) the very thing it was required to show. a = ab + a 2 . Proposition 4" 1 " If a straight-line is cut at random then the square on the whole (straight-line) is equal to the (sum of the) squares on the pieces (of the straight-line), and twice the 52 STOIXEIQN p\ ELEMENTS BOOK 2 ytovicp. A r b rectangle contained by the pieces. A C e H K A Z E EuiMa yap ypamif] f) AB xex^ifjadco, cbc; exu)(ev, xaxd to T. Xeyco, oxi to dira xfjc AB xexpdycovov laov eaxl xoTc xe dno xcov Ar, TB xexpaycovoic xai xcp Sic; Otto xcov Ar, TB Ttepie/o|jievcp opOoycovico. Avayeypdcp'dco yap aTto xfjc AB xexpdycovov xo AAEB, xai CTieCeux^co BA, xal 8id ^ev xou T oTioxepa xcov AA, EB TtapdXX/jXoc fixiSco ^ TZ, 8id 8e xou H OTioxepa xcov AB, AE TiapdXXiqXoc fjy^co f] 0K. xal insl TiapdXXiqXoc eaxiv f] TZ xfj AA, xal eic auxdc e^iTieTixcoxev f) BA, f] exxoc ycovia f) Otto THB i'ar) eaxl xfj evxoc xal dnevavxbv xfj Otto AAB. dXX' f) utto AAB xfj Otto ABA eaxiv lot], snel xal TiXeupd f] BA xfj A A eaxiv lar)' xal f) Otto THB dpa ycovid xfj Otto HBr eaxiv larj- uoie xal TiXeupd f] BT TtXeupa xfj TH eaxiv iar)- dXX' fj [iev TB xfj HK eaxiv for). f\ 5e TH xfj KB- xal f) HK dpa xfj KB eaxiv for)' iaoTtXeupov dpa eaxl xo THKB. Xeyco Br), oxi xal 6pi!)oywviov. etcei yap TiapdXXr)X6c eaxiv f] TH xfj BK [xal eic auxdc e^TieTcxcoxev eu$eTa rj TB] , ai dpa Otto KBr, HrB ycoviai 8uo op'daic eiaiv iaai. opiDf] 8e f] Otco KBE op'df] dpa xal f] Otio BITE coaxe xal ai dnevavxiov ai Otto THK, HKB opiJai eiaiv. op-doycoviov dpa eaxl xo THKB' e8eix'dT) Be xal iaouXeupov xexpdycovov dpa eaxiv xa( eaxiv and xfjc TB. Sid xd aOxd Sf] xal xo OZ xexpdycovov eaxiv xai eaxiv dico xfjc; 6H, xouxeaxiv [dico] xfjc; Ar- xd dpa 0Z, Kr xexpdycova duo xcov Ar, TB eiaiv. xal eicel laov eaxl xo AH xcp HE, xai eaxi xo AH xo Otto xcov Ar, TB- for] yap t\ Hr xfj TB- xal xo HE dpa laov eaxl iu Otto Ar, TB - xd dpa AH, HE iaa eaxl xcp 81c; Otto xcov AF, TB. eaxi Be xal xd OZ, TK xexpdycova djco xcov AF, TB- xd dpa xeaaapa xd 9Z, TK, AH, HE iaa eaxl xolc xe duo xcov Ar, TB xexpaycovoic; xal xco Sic Otio xcov Ar, TB Ttepie)(ofievcp opiJoycovicp. dXXd xd 8Z, TK, AH, HE oXov eaxl xo AAEB, o eaxiv dico xfjc; AB xexpdycovov xo dpa duo xfjc; AB xexpdycovov laov eaxl xolc; xe and xcov Ar, TB xexpaycovoic; xal xcp Sic utto xcov Ar, TB Ttepie)(o(!evcp opiJoycovicp. Edv dpa euiJeTa ypa^f] x^irj'dfj, cbc; exu^ev, xo aTto xfjc For let the straight-line AB have been cut, at random, at (point) C. I say that the square on AB is equal to the (sum of the) squares on AC and CB, and twice the rectangle contained by AC and CB. For let the square ADEB have been described on AB [Prop. 1.46], and let BD have been joined, and let CF have been drawn through C, parallel to either of AD or EB [Prop. 1.31], and let HK have been drawn through G, parallel to either of AB or DE [Prop. 1.31]. And since CF is parallel to AD, and BD has fallen across them, the external angle CGB is equal to the internal and opposite (angle) ADB [Prop. 1.29]. But, ADB is equal to ABD, since the side BA is also equal to AD [Prop. 1.5]. Thus, angle CGB is also equal to GBC. So the side BC is equal to the side CG [Prop. 1.6]. But, CB is equal to GK, and CG to KB [Prop. 1.34]. Thus, GK is also equal to KB. Thus, CGKB is equilateral. So I say that (it is) also right-angled. For since CG is parallel to BK [and the straight-line CB has fallen across them], the angles KBC and GCB are thus equal to two right-angles [Prop. 1.29]. But KBC (is) a right-angle. Thus, BCG (is) also a right- angle. So the opposite (angles) CGK and GKB are also right-angles [Prop. 1.34]. Thus, CGKB is right-angled. And it was also shown (to be) equilateral. Thus, it is a square. And it is on CB. So, for the same (reasons), HF is also a square. And it is on HG, that is to say [on] AC [Prop. 1.34]. Thus, the squares HF and KC are on AC and CB (respectively). And the (rectangle) AG is equal to the (rectangle) GE [Prop. 1.43]. And AG is the (rectangle contained) by AC and CB. For GC (is) equal to CB. Thus, GE is also equal to the (rectangle contained) by AC and CB. Thus, the (rectangles) AG and GE are equal to twice the (rectangle contained) by AC and CB. And HF and GK are the squares on AC and CB (respectively). Thus, the four (figures) HF, GK, AG, and GE are equal to the (sum of the) squares on 53 STOIXEIQN p\ ELEMENTS BOOK 2 okr\z xexpdytovov iaov eaxl xou; xe duo x65v x[iT]|i.dxGJv xe- xpaywvoic; xal xfi 81? utio twv x^ir)[idx«v Tiepiexo^evw op-do- yoviw- oTiep e5ei 8eTc;ai. ^4C and BC, and twice the rectangle contained by AC and CB. But, the (figures) HF, CK, AG, and GE are (equivalent to) the whole of ADEB, which is the square on AB. Thus, the square on AB is equal to the (sum of the) squares on AC and CB, and twice the rectangle contained by AC and CB. Thus, if a straight-line is cut at random then the square on the whole (straight-line) is equal to the (sum of the) squares on the pieces (of the straight-line), and twice the rectangle contained by the pieces. (Which is) the very thing it was required to show t This proposition is a geometric version of the algebraic identity: (a + b) 2 = a 2 + b 2 + 2 a b. 'Edv eOifteTa ypa^r] "W^fi £ ^ ^ Ga xal dviaa, to utio twv dviatov xfjc; okr\c x^i7)^dx«v Tiepie)(o^.evov opiDoyoviov jiexa xoO aTio xfjc ^sia^u xfiv xo[i«v xexpaywvou iaov eaxl xw dTio xfjc f]^iaeiac; xexpayd>v<p. EH Z Eu'deTa yap Tl< ^ ^ AB xex^fja'dto sic, fiev Taa xaxd xo T, sic, 8e aviaa xaxd xo A- Xeyw, oxi xo utio xfiv AA, AB Tiepie)(6[ievov 6p$oya>viov (jiexa xou duo xfjc; TA xexpaywvou iaov eaxl xo duo xfjc; TB xexpaycovco. Avayeypdtfdw yap aTio xfjc; TB xexpdywvov xo TEZB, xal £KeC£U)fdo f) BE, xal 8ia ^tev xoO A oTioxepa iwv TE, BZ TiapdXXrjXoc; f] AH, 5id Se xoO 9 OTioxepa xSv AB, EZ TiapdXXr)Xo<; TidXiv ^/-dco f) KM, xal TtdXiv 8ia xoO A OTioxepa xov TA, BM TiapdXXr]Xoc; fix^w f] AK. xal CTieliaov eaxl xo T0 TiapaTiXf|pco^a xo 0Z napaTiXrjpw^axi, xoivov Tipoaxeia'dw xo AM- oXov dpa xo TM oXw iu AZ 'iaov eaxiv. dXXd xo TM iu AA Iaov eaxiv, ctieI xal f] Ar xfj TB eaxiv Xor\ % xal xo AA dpa xw AZ iaov eaxiv. xoivov Ttpoaxeia'dw xo T0- oXov dpa xo A0 xo MNS^ yvtojxovi Iaov eaxiv. dXXd xo A6 xo utio xwv AA, AB eaxiv Xar] yap f] A0 xfj AB- xal 6 MNS dpa yvwjiwv I'aoc; eaxl x£> utio AA, AB. xoivov itpoaxeiai9M xo AH, o eaxiv iaov xG drco xfjc; TA- 6 dpa MNS yv«[i«v xal xo AH I'aa eaxl iu utio xGv A A, AB uepiexo^tevw op-fJoycovup xal xw drio xfjc; Proposition 5 J If a straight-line is cut into equal and unequal (pieces) then the rectangle contained by the unequal pieces of the whole (straight-line), plus the square on the (difference) between the (equal and unequal) pieces, is equal to the square on half (of the straight-line). A CD B EG F For let any straight-line AB have been cut — equally at C, and unequally at D. I say that the rectangle contained by AD and DB, plus the square on CD, is equal to the square on CB. For let the square CEFB have been described on CB [Prop. 1.46], and let BE have been joined, and let DG have been drawn through D, parallel to either of CE or BE [Prop. 1.31], and again let KM have been drawn through H, parallel to either of AB or EF [Prop. 1.31], and again let AK have been drawn through A, parallel to either of CL or BM [Prop. 1.31]. And since the comple- ment CH is equal to the complement HF [Prop. 1.43], let the (square) DM have been added to both. Thus, the whole (rectangle) CM is equal to the whole (rect- angle) DF. But, (rectangle) CM is equal to (rectangle) AL, since AC is also equal to CB [Prop. 1.36]. Thus, (rectangle) AL is also equal to (rectangle) DF. Let (rect- angle) CH have been added to both. Thus, the whole (rectangle) AH is equal to the gnomon NOP. But, AH 54 STOIXEIQN p\ ELEMENTS BOOK 2 TA xexpaycovo. dXXd 6 MNS yva>(iov xal to AH oXov eaxl to TEZB xexpdywvov, 6 eaxiv duo xrjc TB- to dpa bub tuv AA, AB Tiepiexo^iEvov op^oycoviov jxexa xou diio xrjc TA xexpaycivou laov eaxl xw arco xrjc; TB xeTpaywvw. 'Edv dpa eO'deux ypa^r] xji.Tj'dfj zlc, laa xal aviaa, xo Otco xCSv dviaov xrjc; SXrjc; xur^dxtov Ttepie)(6^evov 6pi9oyMviov ^.exd xoO anb xrjc; (iexac;u xwv xo^iwv xexpaytovou laov eaxl xo duo xrjc; f)(Jioe(ac; xexpayc5va>. onep eBei BeT^ai. is the (rectangle contained) by AD and For DH (is) equal to D.B. Thus, the gnomon NOP is also equal to the (rectangle contained) by AD and DB. Let LG, which is equal to the (square) on CD, have been added to both. Thus, the gnomon NOP and the (square) LG are equal to the rectangle contained by AD and DB, and the square on CD. But, the gnomon NOP and the (square) LG is (equivalent to) the whole square CEFB, which is on CB. Thus, the rectangle contained by AD and DB, plus the square on CD, is equal to the square on CB. Thus, if a straight-line is cut into equal and unequal (pieces) then the rectangle contained by the unequal pieces of the whole (straight-line), plus the square on the (difference) between the (equal and unequal) pieces, is equal to the square on half (of the straight-line). (Which is) the very thing it was required to show. t Note the (presumably mistaken) double use of the label M in the Greek text. t This proposition is a geometric version of the algebraic identity: ab + [(a + 6)/2 - b] 2 = [(a + b)/2] 2 . f '. 'Edv eui9eTa ypa^r] "W^fi <^X a ; TCpoaxeiSfj 8s xu; auxfj eu'deTa in eCWteiat;, xo bnb xrjc; oXrjc; auv xfj Ttpoaxei^tevr] xal xrjc 7tpoaxei[iivr)<; Tiepiexojievov op-fJoytoviov ^texd xoO duo xrjc; f\\u.Gsiac, xexpaycjvou '(gov eaxl x£> duo xrjc; auyxei(ievr)<; ex xe xrjc; fpiaeiac; xal xrjc; 7ipoaxei^evr)<; xexpaywvcp. A r B A E HZ EO'deTa yap xic f] AB xexjirja'dw 8(/a xaxd xo T orp-siov, Tipoaxeia'dw Be xic auxfj euiSela in eCWteiac f] BA- Xeyw, oxi xo utto xaiv AA, AB uepie/ojievov opiSoywviov [icxd xou anb xrjc; TB xexpaywvou I'aov eaxl xfi and xrjc; IA xe- xpaywvw. Avayeypdcp'dw yap anb xrjc; TA xexpdywvov xo TEZA, xal eTiec^eux'Sw rj AE, xal Bid jiev xou B arjjieiou onoxepa xSv Er, AZ TtapdXXiqXoc fj/iSw f] BH, Bid Be xou O arjjieiou onoxepa x£>v AB, EZ TTapdXXrjXoc fj/iSw f) KM, xal exi Bid xou A onoxepa xfiv TA, AM TiapdXXrjXoc fi/iSw f] AK. 'End ouv Tar) eaxlv f) Ar xfj TB, I'aov eaxl xal xo AA Proposition 6+ If a straight-line is cut in half, and any straight-line added to it straight-on, then the rectangle contained by the whole (straight-line) with the (straight-line) having being added, and the (straight-line) having being added, plus the square on half (of the original straight-line), is equal to the square on the sum of half (of the original straight-line) and the (straight-line) having been added. A C B ] D M .0/ / \ / i K L N / / P / / / E G F For let any straight-line AB have been cut in half at point C, and let any straight-line BD have been added to it straight-on. I say that the rectangle contained by AD and DB, plus the square on CB, is equal to the square on CD. For let the square CEFD have been described on CD [Prop. 1.46], and let DE have been joined, and let BG have been drawn through point B, parallel to either of EC or DF [Prop. 1.31], and let KM have been drawn through point H, parallel to either of AB or EF [Prop. 1.31], and finally let AK have been drawn 55 STOIXEIQN p\ ELEMENTS BOOK 2 tu rO. dXXd to TO to ©Z faov eaxiv. xal to AA dpa iu 6Z eaxiv faov. xoivov Tipoaxeiado to TM- oXov apa to AM to NSO yvo^tovi eaxiv faov. dXXd to AM soti to utio tov AA, AB- far) yap ecrav f] AM xfj AB- xal 6 NSO dpa yvo^tov faoc taxi xo utio xov A A, AB [Tiepiex°^ v( P opiJo- yovio]. xoivov Tipoaxdtxdo to AH, 6 saxiv faov xo aTio xfjc Br xsxpayovo- to dpa utio xov AA, AB Tiepie)(6jj£vov opi&oyoviov [lstol tou dno xfjc TB xsxpayovou faov laxl xo NSO yvo^ovi xal ifi AH. dXXd 6 NSO yvojiov xal to AH oXov eraxl to TEZA xexpdyovov, o eaxt,v aTio xfjc TA- to dpa utio xov AA, AB nepi.exo\ievov op-doyoviov ^texd tou arco xfjt; TB xexpayovou faov eaxl to and xfjc FA xexpayovo. 'Edv dpa eu'dsla ypa^if] Tji.rj'dfj Btya, Tipoaxe'dfj 8s tic auxfj eu'dsla etc' eu'deiac, to utio Tfjt; oXr)c auv xfj Ttpo- ox£\.\±evr] xal xfjc Ttpoaxa^iivric Tiepiexo^iEvov op-doyoviov ^.exd xou arco xfjt; fjfiiaeiac xexpayovou faov eaxl to duo Tfjt; auyxei^evr)c ex xe xfjc f)fiiae[ac xal Tfjc Tipoaxei^evr]c xexpayovo- ojcep e5ei Sel^ai. through A, parallel to either of CL or DM [Prop. 1.31]. Therefore, since AC is equal to CB, (rectangle) AL is also equal to (rectangle) CH [Prop. 1.36]. But, (rectan- gle) CH is equal to (rectangle) HF [Prop. 1.43]. Thus, (rectangle) AL is also equal to (rectangle) HF. Let (rect- angle) CM have been added to both. Thus, the whole (rectangle) AM is equal to the gnomon NOP. But, AM is the (rectangle contained) by AD and DB. For DM is equal to DB. Thus, gnomon NOP is also equal to the [rectangle contained] by AD and DB. Let LG, which is equal to the square on BC, have been added to both. Thus, the rectangle contained by AD and DB, plus the square on CB, is equal to the gnomon NOP and the (square) LG. But the gnomon NOP and the (square) LG is (equivalent to) the whole square CEFD, which is on CD. Thus, the rectangle contained by AD and DB, plus the square on CB, is equal to the square on CD. Thus, if a straight-line is cut in half, and any straight- line added to it straight-on, then the rectangle contained by the whole (straight-line) with the (straight-line) hav- ing being added, and the (straight-line) having being added, plus the square on half (of the original straight- line), is equal to the square on the sum of half (of the original straight-line) and the (straight-line) having been added. (Which is) the very thing it was required to show. t This proposition is a geometric version of the algebraic identity: (2 a + b) b + a 2 = {a + b) 2 . 'Edv eu'dsla ypa^tuf] Turj-dfj, oc exu)(ev, to drco Tfjc oXrjc xal to dtp' evoc xov xur^dxov xd auvajicpoxepa xexpdyova faa ecm to xe 81c utio Tfjc oXrjc xal tou eipr^evou xuf]^axoc Tiepiexo[ievo opifayovio xal to arco tou XoitioO xuf^axoc xexpayovo. a r b / / 1 1 A / N / /\ / \ K / / M H / A N E EuiDeTa yap tic rj AB xexjirfa'do, oc exu)(ev, xaxd to F orj^eibv Xeyo, oxi Ta duo xov AB, Br xexpdyova faa eaxl to xe 8lc utio tov AB, Br Tiepiexo^tEvo opifayovio xal to Proposition 7* If a straight-line is cut at random then the sum of the squares on the whole (straight-line), and one of the pieces (of the straight-line), is equal to twice the rectan- gle contained by the whole, and the said piece, and the square on the remaining piece. H / I 1 L / K / / M G D N For let any straight-line AB have been cut, at random, at point C. I say that the (sum of the) squares on AB and BC is equal to twice the rectangle contained by AB and 5G STOIXEIQN p\ ELEMENTS BOOK 2 &7i6 Tfjc; TA TETpayGvo). Avay£ypdcpiL>Gj yap a7l:o T ^ AB TCTpdyiovov to AAEB- xal xaTayeypdcp^co TO a X^ a - 'End ouv Taov eaxl to AH tG HE, xolvov TtpoaxeicrdM to TZ- oXov apa to AZ oXco tG TE Taov eotiv Ta apa AZ, TE SmXaaia soti tou AZ. dXXa Ta AZ, TE 6 KAM taxi yvGjicov xal to rZ TETpdycovov 6 KAM apa yvGjiGJv xal to rZ SiTtXdaid eoti tou AZ. sgti 8e tou AZ SiTtXdaiov xal to 81? utto tGv AB, Br- Tar) yap f) BZ Tfj Br- 6 apa KAM yvGjjiiov xal to TZ TSTpdytovov iaov eoti tG 81c; utio tGv AB, Br. xoivov Tipoaxeio'dw to AH, 6 sgtiv duo Tfjc; Ar TCTpdyiovov 6 apa KAM yvG^tov tal Ta BH, HA TSTpdywva Taa sotl tG ts 61c; utio tGv AB, Br Ttepi£)(ojj£v« op^oycovlcp xal tG arco t^c; Ar TETpayGvip. dXXa 6 KAM yvco^cov xal Ta BH, HA TSTpdytova oXov sotI to AAEB xal to rZ, a eoTiv duo tGv AB, Br TSTpdywva- Ta apa duo xwv AB, Br TETpdywva laa sotI tG [ts] 51c; utio tGv AB, Br TterpiEX ^" 6p{>oywviw ^tSTa tou and Tfjc Ar TSTpayGvou. 'Edv apa euifkTa ypa^jif] T[ir] , df), Gc; stu)(£v, to duo Tfjc; 6Xr)<; xal to dtp' evoc; tGv T^ir^dTtov Ta auva^tcpoTspa TETpdywva Xaa scttI iu ts 81c; utio Tfjc; oXf]z xal tou eipm&vou T[ifj^i.aToc; Ttepie)(o^iv« op'doywvlw xal tG duo tou XoitioO TjifjpcToc; TSTpayGvcp- ouep eSei Belial.. BC, and the square on CA. For let the square ADEB have been described on AB [Prop. 1.46], and let the (rest of) the figure have been drawn. Therefore, since (rectangle) AG is equal to (rectan- gle) GE [Prop. 1.43], let the (square) CF have been added to both. Thus, the whole (rectangle) AF is equal to the whole (rectangle) CE. Thus, (rectangle) AF plus (rectangle) CE is double (rectangle) AF. But, (rectan- gle) AF plus (rectangle) CE is the gnomon KLM, and the square CF. Thus, the gnomon KLM, and the square CF, is double the (rectangle) AF. But double the (rect- angle) AF is also twice the (rectangle contained) by AB and BC. For BF (is) equal to BC. Thus, the gnomon KLM, and the square CF, are equal to twice the (rect- angle contained) by AB and BC. Let DC, which is the square on AC, have been added to both. Thus, the gnomon KLM, and the squares BG and GD, are equal to twice the rectangle contained by AB and BC, and the square on AC. But, the gnomon KLM and the squares BG and GD is (equivalent to) the whole of ADEB and CF, which are the squares on AB and BC (respectively) . Thus, the (sum of the) squares on AB and BC is equal to twice the rectangle contained by AB and BC, and the square on AC. Thus, if a straight-line is cut at random then the sum of the squares on the whole (straight-line), and one of the pieces (of the straight-line), is equal to twice the rect- angle contained by the whole, and the said piece, and the square on the remaining piece. (Which is) the very thing it was required to show. t This proposition is a geometric version of the algebraic identity: (a + b) 2 + a 2 = 2(a + b)a + b 2 . *)'• 'Edv sO'deia ypa^f] T^trydfj, Gc; stu)(£v, to TSTpdxic; Otto Tfjc; bXf]z xal evoc; tGv Tur^aTMv Ttspisxo^evov op-doyGviov ^jLETa tou duo tou Xomou Tuf}pn:oc; TSTpayGvou Taov sgtI tG duo ts Tfjc; bXf]z xal tou eipr}\±evov T^uaToc; Gc; drco ^iiac; dvaypacpevTi TETpayGvw. EO'deTa yap tic; f] AB T£Tjif]ai9M, Gc; stu/sv, xaTa to r ar^elov Xeyw, oti to TSTpdxu; Otto tGv AB, BT rce- pie)(6^.£vov op'doyGviov ^.STa tou duo Tfjc; AT TSTpayGvou iaov so"tI tG duo Tfjc; AB, BT Gc; duo [iiaq dvaypacpevTi TSTpayGvw. 'ExpepXf]OTf)w yap in eG'ddac; [xfj AB eu'dela] f) BA, xal xeioi9« Tfj TB lot) f] BA, xal dvayeypdcp-dw djio Tfjc; A A TSTpdywvov to AEZA, xal xaTayeypdcp'dw SitcXouv to Proposition 8* If a straight-line is cut at random then four times the rectangle contained by the whole (straight-line), and one of the pieces (of the straight-line), plus the square on the remaining piece, is equal to the square described on the whole and the former piece, as on one (complete straight- line) . For let any straight-line AB have been cut, at random, at point C. I say that four times the rectangle contained by AB and BC, plus the square on AC, is equal to the square described on AB and BC, as on one (complete straight-line) . For let BD have been produced in a straight-line [with the straight-line AB], and let BD be made equal to CB [Prop. 1.3], and let the square AEFD have been described on AD [Prop. 1.46], and let the (rest of the) figure have been drawn double. 57 STOIXEIQN p\ ELEMENTS BOOK 2 A r b A B M H 1 \ t<t v i n / Y N O E e A z 'End ouv Tar) eaxlv f] TB xfj BA, dXXd f] (jiev TB xfj HK eaxiv Tar), f] 8e BA xfj KN, xal f] HK apa xfj KN eaxiv Tar). 8id xd auxd Sf) xal f) nP xrj PO eaxiv iar). xal end Iar) eaxlv f] Br XT] BA, f] 8e HK xfj KN, Taov apa eaxl xal xo [lev TK xG KA, xo 8s HP xG PN. dXXd xo TK ifi PN eaxiv Taov napan;Xr]pw|jiaxa yap xou TO 7tapaXXr]Xoypd|jijj.ou- xal xo KA dpa xG HP Taov eaxiv xd xeaaapa dpa xd AK, TK, HP, PN Taa aXXrjXoic; eaxiv. xd xeaaapa apa xexpanXdaid eaxi xou TK. TtdXiv inei iar) eaxlv f) TB xfj BA, dXXd f) [lev BA xrj BK, xouxeaxi xrj TH I'ar), f) 8e TB xfj HK, xouxeaxi xfj Hn, eaxiv iar), xal f) TH apa xfj HEI I'ar) eaxlv. xal excel I'ar) eaxlv f] \xsm TH xfj Hn, f] Se nP xfj PO, Taov eaxl xal xo ^tev AH iw Mn, xo 8e nA xG PZ. dXXd xo Affl xG nA eaxiv Taov TtapaTiXrjpG^axa yap xou MA icapaXXr]Xoypd|i^ou- xal xo AH apa xG PZ Taov eaxiv xd xeaaapa apa xd AH, Mn, nA, PZ Taa dXXf]Xoi<; eaxlv xd xeaaapa apa xou AH eaxi xexpanXdaia. e8d)fdr] 8e xal xd xeaaapa xd TK, KA, HP, PN xou TK xexpaicXdaia- xd apa oxxG, a icepiexei xov ETT yvG^tova, xexpauXdaid eaxi xou AK. xal inei xo AK xo uko xGv AB, BA eaxiv Iar) yap f] BK xfj BA- xo apa xexpdxu; utco xGv AB, BA xexpaicXdaiov eaxi xou AK. eBeix'dr) Se xou AK xexpauXdaioc; xal 6 STT yvGjicov xo apa xexpdxu; utco xGv AB, BA Taov eaxl xG STT yvG^tovi. xoivov Tcpo- axeurdo xo S0, 6 eaxiv Taov xG aTco xrjc Ar xexpayGvor xo dpa xexpdxic; utco xGv AB, BA Tcepie)(6^evov op'doyGviov ^texd xou duo Ar xexpayGvou laov eaxl xG STT yvG^iovi xal xG S0. dXXd 6 STT yvG[i«v xal xo S0 6Xov eaxl xo AEZA xexpdywvov, o eaxiv and xrjc AA - xo apa xexpdxi; utco xGv AB, BA ^texd xou duo Ar Taov eaxl xG and AA xexpayGvcp- Tar] 8e f] BA xfj Br. xo apa xexpdxu; utio xGv AB, Br icepie)(6(ievov op'doyGviov [icxd xou aTco Ar xe- xpayGvou I'aov eaxl xG and xfj; AA, xouxeaxi xG duo xrjc; AB xal Br Gc drco [iidc; dvaypacpevxi xexpayGvw. 'Edv dpa eu'dda ypaji^tf) x^trydfj, G; exu)(ev, xo xexpdxn; utco xfj? okr\z xal evbz xGv x^ir)^dxwv Tiepiexo^ievov opiJoyG- viov ^texd xou dico xou Xoitcou x^tf][iaxoc xexpayGvou laou M O / / G / K \ \ i s / / u Q R/' D N P E H L Therefore, since CB is equal to BD, but CB is equal to GK [Prop. 1.34], and BD to KN [Prop. 1.34], GK is thus also equal to KN. So, for the same (reasons), QR is equal to RP. And since BC is equal to BD, and GK to KN, (square) CK is thus also equal to (square) KD, and (square) GR to (square) RN [Prop. 1.36]. But, (square) CK is equal to (square) RN. For (they are) comple- ments in the parallelogram CP [Prop. 1.43]. Thus, (square) KD is also equal to (square) GR. Thus, the four (squares) DK , CK, GR, and RN are equal to one another. Thus, the four (taken together) are quadruple (square) CK. Again, since CB is equal to BD, but BD (is) equal to BK — that is to say, CG — and CB is equal to GK — that is to say, GQ — CG is thus also equal to GQ. And since CG is equal to GQ, and QR to RP, (rectan- gle) AG is also equal to (rectangle) MQ, and (rectangle) QL to (rectangle) RF [Prop. 1.36]. But, (rectangle) MQ is equal to (rectangle) QL. For (they are) complements in the parallelogram ML [Prop. 1.43]. Thus, (rectangle) AG is also equal to (rectangle) RF. Thus, the four (rect- angles) AG, MQ, QL, and RF are equal to one another. Thus, the four (taken together) are quadruple (rectan- gle) AG. And it was also shown that the four (squares) CK, KD, GR, and RN (taken together are) quadruple (square) CK. Thus, the eight (figures taken together), which comprise the gnomon STU, are quadruple (rect- angle) AK. And since AK is the (rectangle contained) by AB and BD, for BK (is) equal to BD, four times the (rectangle contained) by AB and BD is quadruple (rect- angle) AK. But the gnomon STU was also shown (to be equal to) quadruple (rectangle) AK. Thus, four times the (rectangle contained) by AB and BD is equal to the gnomon STU. Let OH, which is equal to the square on AC, have been added to both. Thus, four times the rect- angle contained by AB and BD, plus the square on AC, is equal to the gnomon STU, and the (square) OH. But, 58 STOIXEIQN p\ ELEMENTS BOOK 2 eaxl xfi dtTio xe xfjc okt]Q xal tou eipr^evou xjarpaxoc; tic, the gnomon STU and the (square) OH is (equivalent to) duo dvaypacpevxi xexpaycovcp- oTiep eSei SeT^ai. the whole square AEFD, which is on AD. Thus, four times the (rectangle contained) by AB and BD, plus the (square) on AC, is equal to the square on AD. And BD (is) equal to BC. Thus, four times the rectangle con- tained by AB and BC, plus the square on AC, is equal to the (square) on AD, that is to say the square described on AB and BC, as on one (complete straight-line). Thus, if a straight-line is cut at random then four times the rectangle contained by the whole (straight-line), and one of the pieces (of the straight-line), plus the square on the remaining piece, is equal to the square described on the whole and the former piece, as on one (complete straight-line) . (Which is) the very thing it was required to show. t This proposition is a geometric version of the algebraic identity: 4 (a + b) a + b 2 = [(a + 6) + a] 2 . •Q'. 'Edv eu'deTa YP°W^ 1 W^Ti ''■ aa xat dviaa, xd dfto x£>v dviawv xfj? oXrjc xur^dxtov xexpdycova SiTiXdaid eoxi xoO xe duo xfjc; r^iadac xod xou dTto xfjc ^exa^u iwv xo^wv xexpaycovou. E A TAB EO'deTa ydp xi? f] AB xex^a'fiw eu; |iev i'aa xaxd xo T, sic, Be aviaa xaxd xo A- Xeyto, oxl xd dno xfiv AA, AB xexpdyiova SiTiXdaid eaxi xfiv dno xGv AT, TA xexpaycovcov. "H/iSio yap duo xou T xfj AB Tipog opiEtdc; f) TE, xal xeicrdio iot) exaxepa xwv AT, TB, xal eTte^eux-dioaav ai EA, EB, xal Sid ^iev xou A xfj ET TiapdXXr]Xoc; rjx$« f) AZ, Bid Be xou Z xrj AB f) ZH, xai £7i£^eu)(i9« f) AZ. xal ensl lay] eaxiv f) Ar xrj TE, iar) eaxl xal i) utio EAT ytovia xfj utio AET. xal eiiel op^rj eaxiv f) Tipoc; x« T, Xoinal apa ai utio EAT, AET [iia. 6pi9fj I'aai eiaiv xai eiaiv iaar r)[iiaeia apa 6pi9rjc; eaxiv exaxepa x£Sv utio TEA, TAE. 8ia xd auxd 8f) xal exaxepa xwv utio TEB, EBT fjjjiiaeid eaxiv 6pi9f)<;- oXr) apa f) utio AEB 6p®r) eaxiv. xal euel f) utio HEZ f]uiaeid eaxiv 6pi9rj<;, op-Qy] Be f) utio EHZ- lot} ydp eaxi xfj evxoc xal aTtevavxiov xfj utio ETB- XoiTtf] apa f) utio EZH f)|iiaeid eaxiv Proposition Qt If a straight-line is cut into equal and unequal (pieces) then the (sum of the) squares on the unequal pieces of the whole (straight-line) is double the (sum of the) square on half (the straight-line) and (the square) on the (dif- ference) between the (equal and unequal) pieces. E A C D B For let any straight-line AB have been cut — equally at C, and unequally at D. I say that the (sum of the) squares on AD and DB is double the (sum of the squares) on AC and CD. For let CE have been drawn from (point) C, at right- angles to AB [Prop. 1.11], and let it be made equal to each of AC and CB [Prop. 1.3], and let EA and EB have been joined. And let DF have been drawn through (point) D, parallel to EC [Prop. 1.31], and (let) FG (have been drawn) through (point) F, (parallel) to AB [Prop. 1.31]. And let AF have been joined. And since AC is equal to CE, the angle EAC is also equal to the (angle) AEC [Prop. 1.5]. And since the (angle) at C is a right-angle, the (sum of the) remaining angles (of tri- angle AEC), EAC and AEC, is thus equal to one right- 59 STOIXEIQN p\ ELEMENTS BOOK 2 op'dfjc;- lor] dpa [eaxlv] f] uno HEZ ycovia xfj uno EZH- mots xal nXeupa f] EH xfj HZ eaxiv '(or). ndXiv etcei f) npog xS B ywvia f)^iaeid eaxiv opi&fjc;, 6p-&y) 8e f) uno ZAB- lot) yap ndXiv eaxl xfj evxoc; xal dnevavxiov xfj uno ErB- Xomf] apa f| utio BZA f]^uaeid eaxiv op'dfjc;- iar) apa f] npoc; iS B yovia xrj uno AZB- waxe xal nXeupa f] ZA nXeupa xrj AB eaxiv tar), xal enel iar] eaxlv f) Ar xfj TE, iaov eaxl xal xo dno Ar iw duo TE- xa apa duo x£>v Ar, TE xexpdytova BmXdaid eaxi xou dno Ar. xou; 8e dno xwv Ar, TE laov eaxl xo dno xfjc; EA xexpdytovov 6pi9f) yap f] Ono ArE ywvia- xo apa duo xfjc; EA 5mXdaiov eaxi xou duo xfjc; Ar. ndXiv, enel Iar] eaxlv f) EH xfj HZ, 'laov xal xo duo xfj? EH xG duo xfjc; HZ- xa apa duo xwv EH, HZ xexpdywva BmXdaid eaxi xou dno xfjc; HZ xexpaytovou. xou; 8e duo xwv EH, HZ xexpayovou; 'laov eaxl xo duo xfjc EZ xexpdywvov xo apa dno xfjc EZ xexpdywvov BmXdaiov eaxi xou dno xfj? HZ. I'ar) 8e f] HZ xr] TA- xo apa duo xfjc EZ SmXdaiov eaxi xou dno xfjc TA. eaxi Be xal xo dno xfj? EA SmXdaiov xou dno xfjc; Ar- xa apa duo xwv AE, EZ xexpdywva BmXdaid eaxi xwv dno xSv Ar, TA xexpaywvwv. xolc Be dno x£>v AE, EZ I'aov eaxl xo dno xfjc; AZ xexpdyovov op-dfj yap eaxiv f) uno AEZ ywvia- xo apa dno xfjc; AZ xexpdywvov 8mXdaiov eaxi xwv dno xc5v Ar, TA. xo 8e dno xfjc; AZ 1'aa xd dno x«v AA, AZ- op-df) yap f] npoc xo A ywvia- xa apa dno xov AA, AZ BmXdaid eaxi xwv dno xtov Ar, TA xexpaywvwv. Tar) 8e f) AZ xfj AB- xa apa dno xwv AA, AB xexpdywva BmXdaid eaxi iwv dno xwv Ar, TA xexpdywvwv. 'Eav apa eu'dela ypa^jif] x^irydfj eic i'aa xal dviaa, xa dno x£>v dviawv xfjc; oXrjc; x[ir]|jidx«v xexpdywva BmXdaid eaxi xou xe dno xfjc; r^iaeiac xal xou dno xfjc; piexacu xov xo^ifiv xexpaycovou- onep eBei SeT^ai. angle [Prop. 1.32]. And they are equal. Thus, (angles) CEA and CAE are each half a right-angle. So, for the same (reasons), (angles) CEB and EEC are also each half a right-angle. Thus, the whole (angle) AEB is a right-angle. And since GEF is half a right-angle, and EGF (is) a right-angle — for it is equal to the internal and opposite (angle) ECB [Prop. 1.29] — the remaining (an- gle) EFG is thus half a right-angle [Prop. 1.32]. Thus, angle GEF [is] equal to EFG. So the side EG is also equal to the (side) GF [Prop. 1.6]. Again, since the an- gle at B is half a right-angle, and (angle) FDB (is) a right-angle — for again it is equal to the internal and op- posite (angle) ECB [Prop. 1.29] — the remaining (angle) BFD is half a right-angle [Prop. 1.32]. Thus, the angle at B (is) equal to DFB. So the side FD is also equal to the side DB [Prop. 1.6]. And since AC is equal to CE, the (square) on AC (is) also equal to the (square) on CE. Thus, the (sum of the) squares on AC and CE is dou- ble the (square) on AC. And the square on EA is equal to the (sum of the) squares on AC and CE. For angle ACE (is) a right-angle [Prop. 1.47]. Thus, the (square) on EA is double the (square) on AC. Again, since EG is equal to GF, the (square) on EG (is) also equal to the (square) on GF. Thus, the (sum of the squares) on EG and GF is double the square on GF. And the square on EF is equal to the (sum of the) squares on EG and GF [Prop. 1.47]. Thus, the square on EF is double the (square) on GF. And GF (is) equal to CD [Prop. 1.34]. Thus, the (square) on EF is double the (square) on CD . And the (square) on EA is also double the (square) on AC. Thus, the (sum of the) squares on AE and EF is double the (sum of the) squares on AC and CD. And the square on AF is equal to the (sum of the squares) on AE and EF. For the angle AEF is a right-angle [Prop. 1.47]. Thus, the square on AF is double the (sum of the squares) on AC and CD. And the (sum of the squares) on AD and DF (is) equal to the (square) on AF. For the angle at D is a right-angle [Prop. 1.47]. Thus, the (sum of the squares) on AD and DF is double the (sum of the) squares on AC and CD. And DF (is) equal to DB. Thus, the (sum of the) squares on AD and DB is double the (sum of the) squares on AC and CD. Thus, if a straight-line is cut into equal and unequal (pieces) then the (sum of the) squares on the unequal pieces of the whole (straight-line) is double the (sum of the) square on half (the straight-line) and (the square) on the (difference) between the (equal and unequal) pieces. (Which is) the very thing it was required to show. t This proposition is a geometric version of the algebraic identity: a 2 + b 2 = 2[([a + b]/2) 2 + ([a + b]/2 - b) 2 ]. 60 STOIXEIQN p\ ELEMENTS BOOK 2 i . 'Edv eu'de'ia ypa^jif] x[j.r]'df) Bixa, Kpoaxe-dfj 8s xi; auxfj eu-dela etc' eu-deia;, to duo xfj; oX/]; auv xfj Kpoaxei^ievr] xal xo dico xfj; Kpoaxeiuevr); xd auva^icpoxepa xexpdywva SijcXdaid eaxi xou xe dico xfj; f^iaeia; xal xou dico xfj; auy- xei[ievr); ex xe xfj; f]^iaeia; xal xfj; Kpoaxeiuevr]; 6c; and [lid; dvaypacpevxo; xexpaycovou. E Z H Eu'de'ia yap xi; f) AB xex^.f|a , dtL> 8[)(a xaxd xo T, Ttpo- axeia^co Se xi; auxfj eu'de'ia etc' eMeia; f) BA- Xeyw, oxi xd dico xwv AA, AB xexpdyiova SmXaaid eaxi xSv dico iSv AT, TA xexpaywvwv. '"H/iJcd yap duo xou T ar)|jieiou xfj AB Kpo; op-da; i) TE, xal xeicdoy Xar\ exaxepa xwv Ar, TB, xal eite^eu)cd«aav ai EA, EB- xal 5id ^iev xou E xfj AA KapdXXrjXo; fjx$w r) EZ, 8ia 8e xou A xfj TE KapdXXrjXo; fjx'&M i) ZA. xal etceI sic; KapaXXf]Xou; eu'deia; xd; Er, ZA eu'deia xi; eveKeaev f] EZ, ai uico TEZ, EZA dpa §ualv op-ddi; I'aai eiaiv ai apa uko ZEB, EZA 8uo op-dGv eXdaaove; eiaiv ai 8e die' eXaaaovwv fj Suo opiDwv expaXXo^tevai au^iKiKxouaiv ai apa EB, ZA expaXXo^ievai era xd B, A ^ep/] au\i- iceaouvxai. expepXf|aT!)waav xal aujiKiKxexwaav xaxd xo H, xal eTreCeu)fdw r] AH. xal eicel Tar) eaxlv f) Ar xfj TE, Xar] eaxi xal ywvia r] uko EAr xfj uko AEr- xal op-Qy) f) Kpo; iu T- f]^iaeia apa op'dfjc; [eaxiv] exaxepa xwv utco EAr, AEr. Bid xd auxa Sf] xal exaxepa xwv uko TEB, EBr f)^uaeid eaxiv op'dfjc;- opiDf] apa eaxlv f\ uko AEB. xal ctccI f)[iiaeia opiDfj; eaxiv f) uko EBr, f^iaeia apa op'dfjc; xal f] uko ABH. eaxi Se xal f] uko BAH 6pi!)rj- tar) yap eaxi xfj uko ATE- evaXXdc^ yap - Xomf] apa f] uko AHB f^iaeid eaxiv op'dfjc;- f) apa uko AHB xfj uko ABH eaxiv i'ar]- waxe xal KXeupd f] B A KXeupa xfj HA eaxiv Xar\. KaXiv, CKel f) uko EHZ f^iaeid eaxiv opiDfjc;, op'df) 8e f) Kpo; xw Z- i'ar] yap eaxi xfj aKevavxiov xfj Kpo; xw r- Xomf] apa i) uko ZEH f^iaeid eaxiv op'dfj;- Tar] apa f) uko EHZ ywvia xfj uko ZEH- waxe xal KXeupd f] HZ KXeupa xfj EZ eaxiv lot}, xal eKel [Tar] eaxlv f) Er xfj TA], Taov eaxi [xal] xo aKO xfj; Er xexpdywvov xw aKo xfj; TA Proposition 10* If a straight-line is cut in half, and any straight-line added to it straight-on, then the sum of the square on the whole (straight-line) with the (straight-line) having been added, and the (square) on the (straight-line) hav- ing been added, is double the (sum of the square) on half (the straight-line), and the square described on the sum of half (the straight-line) and (straight-line) having been added, as on one (complete straight-line). E F For let any straight-line AB have been cut in half at (point) C, and let any straight-line BD have been added to it straight-on. I say that the (sum of the) squares on AD and DB is double the (sum of the) squares on AC and CD. For let CE have been drawn from point C, at right- angles to AB [Prop. 1.11], and let it be made equal to each of AC and CB [Prop. 1.3], and let EA and EB have been joined. And let EF have been drawn through E, parallel to AD [Prop. 1.31], and let FD have been drawn through D, parallel to CE [Prop. 1.31]. And since some straight-line EF falls across the parallel straight-lines EC and FD, the (internal angles) CEF and EFD are thus equal to two right-angles [Prop. 1.29]. Thus, FEB and EFD are less than two right-angles. And (straight-lines) produced from (internal angles whose sum is) less than two right-angles meet together [Post. 5]. Thus, being pro- duced in the direction of B and D, the (straight-lines) EB and FD will meet. Let them have been produced, and let them meet together at G, and let AG have been joined. And since AC is equal to CE, angle EAC is also equal to (angle) AEC [Prop. 1.5]. And the (angle) at C (is) a right-angle. Thus, EAC and AEC [are] each half a right-angle [Prop. 1.32]. So, for the same (rea- sons), CEB and EBC are also each half a right-angle. Thus, (angle) AEB is a right-angle. And since EBC is half a right-angle, DBG (is) thus also half a right- angle [Prop. 1.15]. And BDG is also a right-angle. For it is equal to DCE. For (they are) alternate (angles) Gl STOIXEIQN p\ ELEMENTS BOOK 2 xExpaycivcp- xd apa (xko xo5v Er, TA xsxpdyova SiTtXdaid saxi toO duo xfjc TA xsxpaycbvou. xou; 8s anb t«v Er, FA Taov sax! xo duo xfjc; EA- xo apa duo xfjc; EA xsxpdywvov 8iTtXdaiov saxi xou arco xfjc; Ar xsxpayiovou. TtdXiv, she! Xar\ sax!v f) ZH xfj EZ, Taov sax! xa! xo djio xfjc ZH x£> duo xfjc ZE- xa apa and xwv HZ, ZE SmXdaid saxi xou arco xfj? EZ. xolc 8s duo xwv HZ, ZE taov sax! xo duo xfjc; EH' xo apa aito xfjc; EH 8i7tXdai6v saxi xou and xfjc; EZ. Tar] 8s f] EZ xfj TA- xo apa duo xfjc; EH xsxpdytovov 8i7tXdai6v saxi xou arco xfjc; TA. £8e()fdir] Ss xa! xo diio xfjc; EA 8i7tXdaiov xou arco xfjc; Ar- xd apa duo xwv AE, EH xsxpdycova 8mXdaid saxi iwv diio iwv Ar, TA xsxpaywvwv. xoTc 8s ano xwv AE, EH xsxpaywvoic Taov eax! xo duo xfjc; AH xsxpdycovov xo apa duo xfjc AH 8i7tXdai6v saxi xwv duo xwv Ar, TA. x£> 8s and xfjc; AH Taa eax! xa duo xwv AA, AH- xa apa arco xwv AA, AH [xexpdywva] 8i7tXdaid saxi xwv anb xtov Ar, EA [xexpay u-vov] . Tar] 8e f] AH xfj AB- xd apa aTio iSv AA, AB [xsxpdyova] 8i7iXdaid saxi xwv duo x£Sv Ar, TA xsxpaywvwv. 'Edv apa sO'dsTa ypa^f] xjirj'dfj 8i)(a, TipoaxsiDfj 8s xic auxfj su'dsTa z% sui!)s(ac, xo arco xfjc; oXr]c auv xfj irpo- axsi^isvr] xa! xo duo xfjc; Ttpoaxsi^svr)c xd auva^icpoxspa xsxpdywva SiTtXdaid saxi xou xs duo xfjc; f](jiiasiac xa! xou d:i:6 xfjc auyxsi^svr]c sx xs xfjc; fjuiasiac xa! xfjc; irpo- oxsi\±evr}z cbc duo [iiclc, dvaypacpsvxoc xsxpayovou- oTtsp s8si 8sle;ai. [Prop. 1.29]. Thus, the remaining (angle) DGB is half a right-angle. Thus, DGB is equal to DBG. So side BD is also equal to side GD [Prop. 1.6]. Again, since EGF is half a right-angle, and the (angle) at F (is) a right-angle, for it is equal to the opposite (angle) at C [Prop. 1.34], the remaining (angle) FEG is thus half a right-angle. Thus, angle EGF (is) equal to FEG. So the side GF is also equal to the side EF [Prop. 1.6]. And since [EC is equal to CA] the square on EC is [also] equal to the square on CA. Thus, the (sum of the) squares on EC and CA is double the square on CA. And the (square) on EA is equal to the (sum of the squares) on EC and CA [Prop. 1.47]. Thus, the square on EA is double the square on AC. Again, since FG is equal to EF, the (square) on FG is also equal to the (square) on F E. Thus, the (sum of the squares) on GF and FE is dou- ble the (square) on EF. And the (square) on EG is equal to the (sum of the squares) on GF and FE [Prop. 1.47]. Thus, the (square) on EG is double the (square) on EF. And EF (is) equal to CD [Prop. 1.34]. Thus, the square on EG is double the (square) on CD. But it was also shown that the (square) on EA (is) double the (square) on AC. Thus, the (sum of the) squares on AE and EG is double the (sum of the) squares on AC and CD. And the square on AG is equal to the (sum of the) squares on AE and EG [Prop. 1.47]. Thus, the (square) on AG is double the (sum of the squares) on AC and CD. And the (sum of the squares) on AD and DC is equal to the (square) on AG [Prop. 1.47]. Thus, the (sum of the) [squares] on AD and DG is double the (sum of the) [squares] on AC and CD. And DG (is) equal to DB. Thus, the (sum of the) [squares] on AD and DB is double the (sum of the) squares on AC and CD. Thus, if a straight-line is cut in half, and any straight- line added to it straight-on, then the sum of the square on the whole (straight-line) with the (straight-line) hav- ing been added, and the (square) on the (straight-line) having been added, is double the (sum of the square) on half (the straight-line), and the square described on the sum of half (the straight-line) and (straight-line) having been added, as on one (complete straight-line). (Which is) the very thing it was required to show. t This proposition is a geometric version of the algebraic identity: (2 a + b) 2 + b 2 = 2 [a 2 + (a + ft) 2 ]. ia'. Proposition 11+ Tf]v 8o$sTaav su-dslav xs^sTv waxs xo utto xfjc; oXrjc xa! To cut a given straight-line such that the rectangle xou sxspou xfiv x^rjjidxMv TtspiS)(6[isvov opiSoytoviov Taov contained by the whole (straight-line), and one of the slvai x« duo xou Xoittou x(ifpaxoc; xsxpaywvw. pieces (of the straight-line), is equal to the square on the remaining piece. G2 STOIXEIQN p\ ELEMENTS BOOK 2 T K A 'Eaxco f] 6o r deiaa euiJela f) AB- BsT 8/) xfjv AB xejisiv waxe to utio xfjc oXrjc; xal tou exepou xGv x^ir^dxtov TC£pi£X ^ evov op'doycoviov Taov e!vai xG cxtio tou XoittoO x^ir^axoc; xexpaytovto. AvayeypdcpiJo yap omo xfjc; AB xexpdytovov xo ABAr, xal xex^f]ai9to f) Ar 8[)(a xaxd xo E ar^eiov, xal STre^sux-dto f] BE, xal SiVjx'S" #] TA era xo Z, xal xdcrdco xfj BE Xor\ f) EZ, xal dvayeypacpi9co arco xfjc AZ xexpdycovov xo Z6, xal 8if])(Tf)w f] H0 era xo K- Xeyco, oxi f] AB xex^Tjxai xaxd xo 6, &>ax£ xo Otto xfiv AB, B6 Tiepiex o K £vov opftoycoviov Taov ttoieiv xtp arco xfjc; AO xexpaytovcp. Tkel yap eru'deTa f) Ar xex^trjxai Si/a xaxd xo E, rcpoaxeixai 8e auxfj f] ZA, xo dpa Otto xcov TZ, ZA tis- piexo^tEvov opiJoyMviov [isxd xou aTio xfjc; AE xexpaywvou laov saxl xo arco xfjc EZ xexpaycovco. iar] 8e f] EZ xfj EB- xo dpa Otto xov TZ, ZA ^.exd xou and xfjc; AE I'aov eaxl xfi duo EB. dXXd xw drco EB I'aa eaxl xd duo x£Sv BA, AE- opftf] yap f) Tipoc; ifi A ytovia- xo dpa utto xfiv TZ, ZA jisxd xou duo xfjc; AE I'aov saxl xolc; arco xcov BA, AE. xoivov dcp/]pr]OTf)w xo duo xfjc; AE- Xoittov dpa xo utto xfiv TZ, ZA Tiepiexo^ievov op'doycoviov laov eaxl iw dico xfjc; AB xsxpaytovto. xai eaxi xo jisv utto xiov TZ, ZA xo ZK- Iar) yap f) AZ xfj ZH- xo 8e arco xfjc; AB xo AA- xo dpa ZK Taov eaxl xtp AA. xoivov dpr)pf|a , dco xo AK- Xoittov dpa xo Z6 xfii 0A laov eaxiv. xa( eraxi xo [lev 0A xo utto xcov AB, B0- iar) yap f] AB xfj BA- xo 8e Z6 xo arco xfjc; A9- xo dpa utto xfiv AB, B6 Ttspis/o^ievov opiSoytoviov I'aov eaxl x£> drco 0A xexpaycovco. H dpa 8oi9eTaa eMela f] AB xexpirjxai xaxd xo coaxe xo Otto xcov AB, B0 Trepiexo^ievov 6pi9oycoviov laov ttoisTv to aTTO xfjc; 0A xexpaywvw- orcep e8ei Tioifjaai. C K D Let AB be the given straight-line. So it is required to cut AB such that the rectangle contained by the whole (straight-line), and one of the pieces (of the straight- line), is equal to the square on the remaining piece. For let the square ABDC have been described on AB [Prop. 1.46], and let AC have been cut in half at point E [Prop. 1.10], and let BE have been joined. And let CA have been drawn through to (point) F, and let EF be made equal to BE [Prop. 1.3]. And let the square FH have been described on AF [Prop. 1.46], and let GH have been drawn through to (point) K. I say that AB has been cut at H such as to make the rectangle contained by AB and BR equal to the square on AH. For since the straight-line AC has been cut in half at E, and FA has been added to it, the rectangle contained by CF and FA, plus the square on AE, is thus equal to the square on EF [Prop. 2.6]. And EF (is) equal to EB. Thus, the (rectangle contained) by CF and FA, plus the (square) on AE, is equal to the (square) on EB. But, the (sum of the squares) on BA and AE is equal to the (square) on EB. For the angle at A (is) a right-angle [Prop. 1.47]. Thus, the (rectangle contained) by CF and FA, plus the (square) on AE, is equal to the (sum of the squares) on BA and AE. Let the square on AE have been subtracted from both. Thus, the remaining rectan- gle contained by CF and FA is equal to the square on AB. And FK is the (rectangle contained) by CF and FA. For AF (is) equal to FG. And AD (is) the (square) on AB. Thus, the (rectangle) FK is equal to the (square) AD. Let (rectangle) AK have been subtracted from both. Thus, the remaining (square) FH is equal to the (rectan- gle) HD. And HD is the (rectangle contained) by AB and BH. For AB (is) equal to BD. And FH (is) the (square) on AH. Thus, the rectangle contained by AB G3 STOIXEIQN p\ ELEMENTS BOOK 2 and BH is equal to the square on HA. Thus, the given straight-line AB has been cut at (point) H such as to make the rectangle contained by AB and BH equal to the square on HA. (Which is) the very thing it was required to do. t This manner of cutting a straight-line — so that the ratio of the whole to the larger piece is equal to the ratio of the larger to the smaller piece — is sometimes called the "Golden Section". IP'. 'Ev xoT; djj.pXuy«v[oi; xpiywvoi; to duo xf)<; xfjv d^pXelav ywviav uuoxeivoua/]; uXeupa; xexpdyiovov jiel^ov eaxi t«v duo xGv xf]v d[ipXeTav ywviav uepiexouafiiv uXeupCSv xexpayGvojv xfi uepiexo^ievcp Su; Ouo xe [iia; xwv uepl xr]v d^pXeTav ycoviav, ecp' f]v f] xdi&exo; uiuxei, xai xfj; duoXajipavo^evT]; exxo; (mo xfj; xa-dexou upo; xfj djipXeia ycovia. B A a r "Eaxco djjipXuYcdviov xpiywvov xo ABr d^pXelav e/ov xr)V Ouo BAr, xai fix^" duo xoO B ar]^efou era xf)v EA expX/j'daaav xdiJexo; f] BA. Xeyw, oxi xo duo xfj; Br xexpdycovov \is%ov eaxi xwv duo xov BA, Ar xexpaycovov xo 51? Ouo xwv EA, AA uepiexo^evcp op-doywviw. 'Euel yap eu'dela f] EA xex|ir]xai, <i>; exuxev, xaxd xo A arjjj.elov, xo dpa duo xfj? Ar 1'aov eaxi xoT; duo xtov EA, AA xexpaycovoi; xai xfi 51? Ouo xfiv EA, AA uepiexojievip 6pi9oya>v[<j. xoivov upoaxeiaiJo xo duo xfj? AB' xd dpa duo x£>v TA, AB I'aa eaxl xoT; xe duo xCSv LA, AA, AB xe- xpaycovoi; xai xtp 61c; Ouo xCSv BA, AA [uepiexo^ievti> opiSo- ycovicp]. dXXd xol; ^tev duo xov TA, AB Taov eaxl xo duo xfj; TB- op-df] yap f) upo; xto A ywvia- xoT? 8e duo xwv A A, AB i'aov xo duo xfj; AB- xo dpa duo xfj; TB xexpdywvov Taov eaxl xoT; xe duo xwv TA, AB xexpaycovoi; xai xco 51; Ouo xcov BA, AA uepiexo^ievcp op^oycovicp- oaxe xo duo xfj; LB xexpdycovov xcov duo xcov TA, AB xexpaywvov ^icTCov eaxi tu 81; Ouo xcov BA, AA uepiexo^evco opiSoycovicp. 'Ev dpa xoT; d^pXuycovioi; xpiycovoi; xo duo xfj; xf]v d^pXelav ycoviav Ouoxeivooar]? uXeopd; xexpdycovov y.s%6\i eaxi ifiv duo xcov xf)V d^tpXeTav ycoviav uepiexooacov Proposition 12+ In obtuse-angled triangles, the square on the side sub- tending the obtuse angle is greater than the (sum of the) squares on the sides containing the obtuse angle by twice the (rectangle) contained by one of the sides around the obtuse angle, to which a perpendicular (straight-line) falls, and the (straight-line) cut off outside (the triangle) by the perpendicular (straight-line) towards the obtuse angle. B DA C Let ABC be an obtuse-angled triangle, having the an- gle BAC obtuse. And let BD be drawn from point B, perpendicular to CA produced [Prop. 1.12]. I say that the square on BC is greater than the (sum of the) squares on BA and AC, by twice the rectangle contained by CA and AD. For since the straight-line CD has been cut, at ran- dom, at point A, the (square) on DC is thus equal to the (sum of the) squares on CA and AD, and twice the rectangle contained by CA and AD [Prop. 2.4]. Let the (square) on DB have been added to both. Thus, the (sum of the squares) on CD and DB is equal to the (sum of the) squares on CA, AD, and DB, and twice the [rect- angle contained] by CA and AD. But, the (square) on CB is equal to the (sum of the squares) on CD and DB. For the angle at D (is) a right-angle [Prop. 1.47]. And the (square) on AB (is) equal to the (sum of the squares) on AD and DB [Prop. 1.47]. Thus, the square on CB is equal to the (sum of the) squares on CA and AB, and twice the rectangle contained by CA and AD. So the square on CB is greater than the (sum of the) squares on G4 STOIXEIQN p\ ELEMENTS BOOK 2 TtXeupfiv xexpaytovwv xfi Ttepixo^iivw 81? Otto xe [iia? xwv CA and AB by twice the rectangle contained by CA and itepl xf]v djipXeTav ywviav, ecp' fjv #] xdiJexo? tuttxcl, xal xfj? AD. dmoXa[ipavo^£VT)<; exxo? bnb xfj? xaiiexou rcpo? xfj d^pXeia Thus, in obtuse-angled triangles, the square on the ycoviqr ojiep eSei 8eTc;ai. side subtending the obtuse angle is greater than the (sum of the) squares on the sides containing the obtuse an- gle by twice the (rectangle) contained by one of the sides around the obtuse angle, to which a perpendicu- lar (straight-line) falls, and the (straight-line) cut off out- side (the triangle) by the perpendicular (straight-line) to- wards the obtuse angle. (Which is) the very thing it was required to show. t This proposition is equivalent to the well-known cosine formula: BC 2 = AB 2 + AC 2 - 2 AB AC cos BAC, since cos BAC = —AD/AB. *Ev xoT? 6c;uyGJv[oi? xpiycjvoi? xo duo xfj? xrjv 6c;eTav ycoviav unoxeivouar]? jtXeupd? xexpdywvov eXaxxov laxi xfiv duo x«v xr)v o^elav yioviav nspis/ouawv TtXeupfiv xe- xpayiovtov xai Ttepiexo(jievG3 81? Otto xe [iia? xfiv Ttepl xfjv 6i;eTav ywviav, ecp' f]v f] xd-dexo? TUTtxei, xdi xrjc; dTtoXajjipa- vojievr)? evxo? Otto xfjc; xai9exou Ttpo? xfj o^eitx y«v(a. A BAT 'Eaxco 6c;uy«viov xpiywvov xo ABr 6c;eTav e^ov xf]v Ttpo? iS B ywviav, xdi /jx^" diio xoO A ar^dou Ira xf]v Br xd-dexo? f\ AA- Xeyco, oxi xo duo xfjc Ar xexpdycovov eXaxxov eaxi xCSv duo xov TB, BA xexpaywvov xc5 Sic; tmb xov TB, BA 7i£piex°^E vt P op-doycoviw. Tkel yap cuiMa f] TB xex^trjxai, d>? exuxev, xaxd xo A, xd dpa omb xfiv TB, BA xexpdycova laa eaxl xw xe 81? Otto xfiv TB, BA Ttepiexo^tevw op'doywvta) xdi xfii duo xfj? Ar xexpaywvo. xoivov upooxeia'dw xo duo xfj? AA xexpdytovov xd dpa duo xwv TB, BA, AA xexpdywva laa eaxl tu xe 81? utio twv TB, BA uepiexo^evw opi^oyoviw xal xoT? dfto xwv AA, Ar xexpaywvio?. dXXa xoT? [lev arto iwv BA, AA I'aov xo duo xfj? AB' op'df) yap f] Ttpo? xw A ycovia- xoT? 8e duo xwv AA, Ar I'aov xo duo xfj? Ar- xd dpa duo xtov TB, BA laa eaxl x£3 xe arco xfj? Ar xal tu 81? utco twv TB, BA- waxe ^tovov xo arto xfj? Ar eXaxxov eaxi Proposition 13+ In acute-angled triangles, the square on the side sub- tending the acute angle is less than the (sum of the) squares on the sides containing the acute angle by twice the (rectangle) contained by one of the sides around the acute angle, to which a perpendicular (straight-line) falls, and the (straight-line) cut off inside (the triangle) by the perpendicular (straight-line) towards the acute angle. A B D C Let ABC be an acute-angled triangle, having the an- gle at (point) B acute. And let AD have been drawn from point A, perpendicular to BC [Prop. 1.12]. I say that the square on AC is less than the (sum of the) squares on CB and BA, by twice the rectangle contained by CB and BD. For since the straight-line CB has been cut, at ran- dom, at (point) D, the (sum of the) squares on CB and BD is thus equal to twice the rectangle contained by CB and BD, and the square on DC [Prop. 2.7]. Let the square on DA have been added to both. Thus, the (sum of the) squares on CB, BD, and DA is equal to twice the rectangle contained by CB and BD, and the (sum of the) squares on AD and DC. But, the (square) on AB (is) equal to the (sum of the squares) on BD and DA. For the angle at (point) D is a right-angle [Prop. 1.47]. G5 STOIXEIQN p\ ELEMENTS BOOK 2 x£5v duo xcov TB, BA xsxpaytovtov xcp 5k uno xfiv TB, BA 7tepiexo[ji£V(p op-doyMviw. 'Ev dpa xok o^uytovbic; xpiyiovoic; xo dmo xrjg xr)v o^slav ywviav UTtoxeivoua/jt; TtXeup&c; xexpdywvov eXaxxov eaxi xwv duo xcov xrjv o^eiav ytoviav rcspiexouatov nXeupfiv xs- xpaycoviov xio Ttspis/o^ievcp 5k 0716 xs [iiac, xfiv Txepl xr]v o^slav ycoviav, ecp' fjv f) xd-dexoc; tutcxsi, xdi xrjc aTtoXa^pa- vojievrjc; svxoc Otco xfjc xaiJexou Ttpoc; xrj o^eia y«v[qc oTtep eBei 5dc;ai. And the (square) on AC (is) equal to the (sum of the squares) on AD and DC [Prop. 1.47]. Thus, the (sum of the squares) on CB and BA is equal to the (square) on AC, and twice the (rectangle contained) by CB and BD. So the (square) on AC alone is less than the (sum of the) squares on CB and BA by twice the rectangle contained by CB and BD. Thus, in acute-angled triangles, the square on the side subtending the acute angle is less than the (sum of the) squares on the sides containing the acute angle by twice the (rectangle) contained by one of the sides around the acute angle, to which a perpendicular (straight-line) falls, and the (straight-line) cut off inside (the triangle) by the perpendicular (straight-line) towards the acute angle. (Which is) the very thing it was required to show. t This proposition is equivalent to the well-known cosine formula: AC 2 = AB 2 + BC 2 — 2 AB BC cos ABC, since cos ABC = BD /AB. r J A 'Eaxto xo 5oi3ev eu-duypa^ov xo A- 5eT 5rj xw A eG'duypd^w 1'aov xexpdywvov auaxrpaaTJai. Suveaxdxw ydp xQ A eu'duypd^iw urov TtapaXXrjXo- ypa^ov op'doywviov xo BA - d \itv ouv iar\ eaxlv f] BE xrj EA, yeyovot; dv sir] xo emxax'dev. auveaxaxai ydp x£> A eu'duypd^w i'aov xexpdywvov xo BA- ei 5e ou, ^iia xfiv BE, EA jiei^cov eaxiv. eaxo ^dCtov f] BE, xdi expepXria'do etc! xo Z, xdi xeiaaJw xrj EA for) t\ EZ, xdi xex^trjo'dw f] BZ 5[)(a xocxd xo H, xdi xevxpa) xw H, 5iaoxf]^axi 5e evl xwv HB, HZ f]^ixuxXiov y£ypdcpif)w xo B9Z, xdi expepXf]a'do f) AE era xo 6, xdi ineZs^x^ ^ HQ. Tkei ouv eO'deia f\ BZ xex^trjxai etc; y.sv foa xaxd xo H, tic, be aviaa xaxd xo E, xo dpa Oreo xfiv BE, EZ 7i£pi£x°HE vov opiDoywviov ^.exd xou duo xfj<; EH xexpaycovou i'aov eaxi tu aTto xfjc HZ xexpaywvw. for] 5s f] HZ xrj HQ- xo dpa U7i6 xwv BE, EZ ^texd xoO aito xrj<; HE i'aov lax! iw dfto xfjc; HQ. iS 5e duo xfj<; HQ foa eaxi xd duo xwv QE, EH D Let A be the given rectilinear figure. So it is required to construct a square equal to the rectilinear figure A. For let the right-angled parallelogram BD, equal to the rectilinear figure A, have been constructed [Prop. 1.45]. Therefore, if BE is equal to ED then that (which) was prescribed has taken place. For the square BD, equal to the rectilinear figure A, has been constructed. And if not, then one of the (straight-lines) BE or ED is greater (than the other). Let BE be greater, and let it have been produced to F, and let EF be made equal to ED [Prop. 1.3]. And let BF have been cut in half at (point) G [Prop. 1.10]. And, with center G, and radius one of the (straight-lines) GB or GF, let the semi-circle BHF have been drawn. And let DE have been produced to H, and let GH have been joined. Therefore, since the straight-line BF has been cut — equally at G, and unequally at E — the rectangle con- 66 STOIXEIQN p\ ELEMENTS BOOK 2 xexpdywva - to apa utto tSv BE, EZ \iexa. xou duo HE laa eaxl xoT<; aito xfiv 8E, EH. xoivov dcp/jprjo-dcL) to duo xfj<; HE xexpdywvov Xomov apa to (mo xov BE, EZ Tiepiexo^evov opiSoycjviov I'aov saxl xai duo xrjg EG xsxpaytovto. dXXa xo utto xfiv BE, EZ xo BA eaxiv Xar] yap f) EZ xfj EA- xo apa BA 7iapaXXr)X6ypa[i^ov iaov soxl xG duo xfjc; 0E xe- xpaytovtp. 1'aov 8s xo BA x£i A Euduypdji^co. xal xo A apa £ui36ypa|ji[j.ov i'aov sax! xcp duo xfjc EO dvaypacprjao^evw xexpaycivw. Tw apa So'devxi eu'duypd^cp x£> A laov xexpdywvov auveaxaxai xo duo xfjc E6 dvaypacp^ao^ievov onep eSei noifjaai. tained by BE and SF, plus the square on EG, is thus equal to the square on GF [Prop. 2.5]. And GF (is) equal to GH. Thus, the (rectangle contained) by BE and EF, plus the (square) on GE, is equal to the (square) on GH. And the (sum of the) squares on HE and EG is equal to the (square) on GH [Prop. 1.47]. Thus, the (rectangle contained) by BE and EF, plus the (square) on GE, is equal to the (sum of the squares) on HE and EG. Let the square on GE have been taken from both. Thus, the remaining rectangle contained by BE and EF is equal to the square on EH. But, BD is the (rectangle contained) by BE and EF. For EF (is) equal to ED. Thus, the par- allelogram BD is equal to the square on HE. And BD (is) equal to the rectilinear figure A. Thus, the rectilin- ear figure A is also equal to the square (which) can be described on EH. Thus, a square — (namely), that (which) can be de- scribed on EH — has been constructed, equal to the given rectilinear figure A. (Which is) the very thing it was re- quired to do. 67 68 ELEMENTS BOOK 3 Fundamentals of Plane Geometry Involving Circles 69 ETOIXEIfiN y'. ELEMENTS BOOK 3 "Opoi. a', 'laoi xuxXoi Eiaiv, Gv ai 8id|iExpoi i'aai siaiv, fj Gv ai ex iwv xsvxpov laai eiaiv. P'. Eu-dsTa xuxXou sqxxTCXEcrdai XsYSxai, rjxu; aTCxo^isvr) xou xuxXou xal sxPaXXo^isvr] ou xe^ivei xov xuxXov. y'. KuxXoi scpdTCxscrdai dXXrjXtov Xcyoviai o'lxivsc; aTixo- ^.svoi dXXrjXwv ou xsjivouaiv dXXiqXouc;. 8'. 'Ev xuxX« i'aov ditExsiv duo xou xsvxpou su-dslai Xsyovxai, oxav ai duo xoO xsvxpou etc'' auxdc xadsxoi dyo^ievai '(aai Gaiv. e'. MeTCov 8e aTCExsiv XEYSxai, Icp' fjv f] ^isiCwv xdi9sxo<; TUTCXEl. 9'. T^ifj^a xuxXou lax! xo TCEpisxo^tEvov axrj(ia utco xs eu-Qsiac, xal xuxXou TCEpicpspsiac;. T^r|uaxo<; 8s ywvia saxiv f] TCspisxo^isvr] utco xs su'dsia? xal xuxXou TCEpicpspsiac;. T)'. 'Ev x^irj^iaxi 8s yavia saxiv, oxav era xfj? TCspi- cpspsiac; xou x^rj^axoe; Xrjcp'dfj xi ar)(ieTov xal octc' auxou era xa nspaxa xrj<; su'dsia?, fj saxi pdau; xou x^ir^axoc;, stcl- ^eux^woiv su-dsla!,, f] TCepiexo^Evr] ywvia utco xGv stci^su- X'dsiaGv sutJeiGv. "Oxav 8e ai Ttspisxouaai x/]v ywviav su'dslai aito- Xa^pdvoal xiva TCEpicpspsiav, etc' exei\>r\q Xsysxai psprjXEvai f] Ywvia. 1'. To^isix; 8s xuxXou saxiv, oxav Ttpo<; xG xsvxpG xou xuxXou auaxa-dfj Y^via, xo TCEpisxo^svov axrjjia utco xe iSv x/]v Y"v(av TiEpiExouawv su-dsiGv xal xfjc aTCoXa^pavo^svr](; utc' auxGv TCspicpspsiac;. ia'. "O^ioia x[if]^axa xuxXwv Eaxi xa Ssxo^isva ycoviat; Iaa<;, fj ev olz ai Y^viai laai dXXr]Xai<; siaiv. a . Tou 8oi9svxo<; xuxXou xo xsvxpov supslv. 'Eax« 6 So-dsli; xuxXoc 6 ABE 5sT 8rj xou ABr xuxXou xo xsvxpov supsTv. Air|x$w xu; si<; auxov, G<; sxuxsv, su'dsTa f) AB, xal x£x^tr]OTf)w 8ixa xaxa xo A ar\\ieiov, xal aito xou A xfj AB 7ipo<; 6pM<; rjx'dw f] Ar xal Sifix^w etc! xo E, xal xsxuf|a , d« f] TE 8ixa xaxa xo Z- Xsy«, oxi xo Z xsvxpov saxi xou ABr [xuxXou] . Mf] yap, dXX' si 5uvax6v, saxw xo H, xal ETiECsux'dwCTav ai HA, HA, HB. xal etcei for) saxlv ?) AA xrj AB, xoivr) 8e f) AH, 860 8/] ai AA, AH 860 xdi<; HA, AB i'aai siaiv sxaxspa sxaxspa- xal pdau; f) HA pdasi xrj HB iaxiv tar)- ex xsvxpou Yap- Y«via apa f) utco AAH Y^via xfj utco HAB iaf] saxiv. Definitions 1. Equal circles are (circles) whose diameters are equal, or whose (distances) from the centers (to the cir- cumferences) are equal (i.e., whose radii are equal). 2. A straight-line said to touch a circle is any (straight- line) which, meeting the circle and being produced, does not cut the circle. 3. Circles said to touch one another are any (circles) which, meeting one another, do not cut one another. 4. In a circle, straight-lines are said to be equally far from the center when the perpendiculars drawn to them from the center are equal. 5. And (that straight-line) is said to be further (from the center) on which the greater perpendicular falls (from the center). 6. A segment of a circle is the figure contained by a straight-line and a circumference of a circle. 7. And the angle of a segment is that contained by a straight-line and a circumference of a circle. 8. And the angle in a segment is the angle contained by the joined straight-lines, when any point is taken on the circumference of a segment, and straight-lines are joined from it to the ends of the straight-line which is the base of the segment. 9. And when the straight-lines containing an angle cut off some circumference, the angle is said to stand upon that (circumference) . 10. And a sector of a circle is the figure contained by the straight-lines surrounding an angle, and the circum- ference cut off by them, when the angle is constructed at the center of a circle. 11. Similar segments of circles are those accepting equal angles, or in which the angles are equal to one an- other. Proposition 1 To find the center of a given circle. Let ABC be the given circle. So it is required to find the center of circle ABC. Let some straight-line AB have been drawn through (ABC), at random, and let (AB) have been cut in half at point D [Prop. 1.9]. And let DC have been drawn from D, at right-angles to AB [Prop. 1.11]. And let (CD) have been drawn through to E. And let CE have been cut in half at F [Prop. 1.9]. I say that (point) F is the center of the [circle] ABC. For (if) not then, if possible, let G (be the center of the circle), and let GA, CD, and GB have been joined. And since AD is equal to DB, and DG (is) common, the two 70 ETOIXEIfiN y'. ELEMENTS BOOK 3 oxav 8e euif)eTa in euiJeTav axaiSeTaa xdg ecpe<;rj<; ytovtag laac, dXXrjXaic; ranfj, dpfti) exaxepa xaiv lacov yioviGv eaxiv 6p0r] dpa eaxlv f) Otto HAB. eaxl Be xal f] utto ZAB opTEby Tor) apa f) Otto ZAB xfj Otto HAB, f] [ici^mv xrj eXdxxovr onep eaxlv dBuvaxov. oux dpa xo H xevxpov eaxl xou ABr xuxXou. by.o'\.Gic, Br] Be^ojiev, oxi ouB' dXXo xi TtXf]v xoO Z. r H z \ A (straight-lines) AD, DC are equal to the two (straight- lines) BD, DG^ respectively. And the base GA is equal to the base GB. For (they are both) radii. Thus, angle ADG is equal to angle GDB [Prop. 1.8]. And when a straight-line stood upon (another) straight-line make ad- jacent angles (which are) equal to one another, each of the equal angles is a right-angle [Def. 1.10]. Thus, GDB is a right-angle. And FDB is also a right-angle. Thus, FDB (is) equal to GDB, the greater to the lesser. The very thing is impossible. Thus, (point) G is not the center of the circle ABC. So, similarly, we can show that neither is any other (point) except F. c F G \ D E To Z dpa arj^elov xevxpov eaxl xou ABT [xuxXou]. Thus, point F is the center of the [circle] ABC. riopiajaa. 'Ex Br) xouxou cpavepov, oxi edv ev xuxXw cu-deld uc, eMeldv xiva 8()(a xal upoc; op-ddc; xejivr), Era xfjc; xejivouarjc; eaxl xo xevxpov xou xuxXou. — ouep eBei icoirjaai. t The Greek text has "GD, DB", which is obviously a mistake. P'- 'Edv xuxXou era xfj? uepicpepeiac; Xrjcp'dfi Buo xu)(6vxa a/jueTa, f] era xd ar^eTa era^euyvu^evr) eO'dela evxoc; rceaelxai xou xuxXou. Tiaxw xuxXo<; 6 ABT, xal era xfj? uepicpepeiac; auxou eiXrjcpiEko Buo xu)(6vxa a/jjiela xd A, B- Xeyto, oxi f] duo xou A era xo B era^euYvu^evrj euiSela evxog jieaelxai xou xuxXou. Mr] y^Pj «XX' el 6uvax6v, raTtxexco exxoc <i>c; fj AEB, xal eiXricp'dw xo xevxpov xou ABT xuxXou, xal eaxw xo A, xal ejceCeux'dwaav al AA, AB, xal 8ir|X$« f] AZE. 'Etcei ouv lar\ eaxlv f] A A xrj AB, iar\ apa xal ytovta f) utco AAE xrj uko ABE- xal excel xpiywvou xou AAE ^iia Corollary So, from this, (it is) manifest that if any straight-line in a circle cuts any (other) straight-line in half, and at right-angles, then the center of the circle is on the for- mer (straight-line) . — (Which is) the very thing it was required to do. Proposition 2 If two points are taken at random on the circumfer- ence of a circle then the straight-line joining the points will fall inside the circle. Let ABC be a circle, and let two points A and B have been taken at random on its circumference. I say that the straight-line joining A to B will fall inside the circle. For (if) not then, if possible, let it fall outside (the circle), like AEB (in the figure). And let the center of the circle ABC have been found [Prop. 3.1], and let it be (at point) D. And let DA and DB have been joined, and let DFE have been drawn through. Therefore, since DA is equal to DB, the angle DAE 71 ETOIXEIfiN y'. ELEMENTS BOOK 3 TiXeupd TtpoaexpepXrjxai f] AEB, [isi^cov apa f\ hub AEB y«v(a xfjc utco AAE. Tar) 8e f] utio AAE xrj (mo ABE- ^tei^wv apa f] tmo AEB xrj<; O716 ABE. utio 8e xf|V ^tel^ova ywviav f) jieii^wv TtXeupd unoxeivei' [iei^iov apa f] AB xfjc; AE. Tar] 8e f] AB xrj AZ. jie^wv apa r) AZ xrjc; AE f) eXdxxcov xrjc; ^iei^ovoc;- ouep eaxlv d8uvaxov. oux apa f] dtTio xou A era xo B eiuCeuyvu^evr] eu'dela exxoc; Tteaelxai xoO xuxXou. o^iolwc; 8r] Se^o^isv, oxi ouBe en' auxrjc; xrjc; Ttepicpepelac;- evxoc; apa. 'Eav apa xuxXou em xfjc; nepicpepeiac; X/]cpif)fj 860 xu)(6vxa a/)[ieTa, f) era xa ar)[ieTa eTti^euyvu^evr) euiJeTa evxoc; Tteaelxai xoO xuxXou- oitep e8ei SeT^ai. Y • 'Eav ev xuxXcp eu^sTa uc, 81a xou xevxpou euiMdv xiva \ir\ 8id xoO xevxpou 8i)(a xejivr], xal Ttpoc; op'dac; auxrjv xejivei- xal eav Tipog opMc; auxr]v xejivr), xal 8(xa aux/jv xejivei. "Eaxa> xuxXoc; 6 ABr, xal ev auxcp eO'deTd xi<; 61a xou xevxpou f) TA eMeldv xiva \ly] Bid xou xevxpou xrjv AB Bi/a xejivexco xaxd xo Z arj^ielov Xeyco, 6x1 xal Ttpoc; 6pi)ac; auxrjv xe^ivei. EiXrjCpiftco yap xo xevxpov xou ABr xuxXou, xal eaxco xo E, xal £K£^eu)(i9«oav al EA, EB. Kod CTtelTar) eaxlv f] AZ xrj ZB, xolvt) Be rj ZE, 860 8ualv I'oai [etatv] 1 xal pdaic; f) EA pdaei xrj EB tor) 1 yiovia apa f) Oito AZE yiovia xrj Otto BZE Tar) eaxiv. oxav Be eu-dela Itc' eu-delav axaOeTaa xdc; ecpec^fjc; ycoviac; Taac dXXfjXaic; Ttoifj, opiDr] exaxepa xwv Tacov ywvifiv eaxiv exaxepa apa xwv UTto AZE, BZE opftfi eaxiv. i] EA apa Sid xou xevxpou ouaa xrjv AB [Li] Bid xou xevxpou ouaav Bixa xepivouaa xal Ttpoc; 6pM<; xejivei. (is) thus also equal to DBE [Prop. 1.5]. And since in tri- angle DAE the one side, AEB, has been produced, an- gle DEB (is) thus greater than DAE [Prop. 1.16]. And DAE (is) equal to DBE [Prop. 1.5]. Thus, DEB (is) greater than DBE. And the greater angle is subtended by the greater side [Prop. 1.19]. Thus, DB (is) greater than DE. And DB (is) equal to DF. Thus, DF (is) greater than DE, the lesser than the greater. The very thing is impossible. Thus, the straight-line joining A to B will not fall outside the circle. So, similarly, we can show that neither (will it fall) on the circumference itself. Thus, (it will fall) inside (the circle) . Thus, if two points are taken at random on the cir- cumference of a circle then the straight-line joining the points will fall inside the circle. (Which is) the very thing it was required to show. Proposition 3 In a circle, if any straight-line through the center cuts in half any straight-line not through the center then it also cuts it at right-angles. And (conversely) if it cuts it at right-angles then it also cuts it in half. Let ABC be a circle, and, within it, let some straight- line through the center, CD, cut in half some straight-line not through the center, AB, at the point F. I say that (CD) also cuts (AB) at right-angles. For let the center of the circle ABC have been found [Prop. 3.1], and let it be (at point) E, and let EA and EB have been joined. And since AF is equal to FB, and FE (is) common, two (sides of triangle AFE) [are] equal to two (sides of triangle BFE). And the base EA (is) equal to the base EB. Thus, angle AFE is equal to angle BFE [Prop. 1.8]. And when a straight-line stood upon (another) straight- line makes adjacent angles (which are) equal to one an- other, each of the equal angles is a right-angle [Def. 1.10]. Thus, AFE and BFE are each right-angles. Thus, the 72 ETOIXEIfiN y'. ELEMENTS BOOK 3 r A AXXd 8f] f] TA xrjv AB icpoc; op'dag xejivexav Xeyw, oxi xal 8ixa auxrjv xe^tvei, xouxeaxiv, oxi I'ar) eaxlv f] AZ xfj ZB. Tt5v yap auxwv xaxaaxeuaai)evxMv, etcei iar\ Eaxlv f) EA xfj EB, iar] eaxl xal ywvia f] utxo EAZ xfj uito EBZ. eaxl Se xal op-dr) rj Otco AZE op'dfj xfj (mo BZE iar)- 860 dpa xpiywvd eaxi EAZ, EZB xa<; 860 ywv(a<; 8uol ywviau; Iaa<; E^ovxa xal jiiav jcXeupdv [iia TcXeupa I'arjv xoivrjv auxfiv xrjv EZ uicoxeivouaav uko ^uav xGv lawv ywviwv xal xa<; XoiKag apa TcXeupac; xau; Xomau; icXeupau; Xaac, e^ei- larj apa rj AZ xfj ZB. 'Eav apa ev xuxXw EU'dsTd uc, 81a xou xevxpou euiMdv xiva [ufj 81a xou xsvxpou Sixa xe^vrj, xal jcpo<; op^a? auxrjv xejuver xal eav Ttpo<; opMc; auxrjv xeuvrj, xal 8[^a auxrjv xe^tvei- oicep eSei BeT^ai. 5'. Eav ev xuxXcp 860 eu-deTai xe^ivcoaiv dXXrjXag [urj 8la xou xevxpou ouaai, ou xs^vouaiv dXXrjXac; 8[)(a. 'Eaxco xuxXo<; 6 ABrA, xal ev auxcp 860 eO'delai ai Ar, BA xe^ivexMoav dXXrjXa<; xaxa xo E [if] Sid xou xevxpou ouaai- Xeyw, oxi ou xejivouaiv dXXf]Xa<; 8[)(a. EE yap Suvaxov, xe^vexoaav dXXrjXac; 8[)(a &axe Tarjv eTvai xrjv |iev AE xrj Er, xrjv Be BE xrj EA- xal eiXrjcpiL>Gj xo xevxpov xou ABrA xuxXou, xal eaxco xo Z, xal ejie^eu)cd« f] ZE. 'EticI ouv eO'deld xig 81a xou xevxpou f] ZE eui&eTdv xiva (if) Bid xou xevxpou xrjv Ar 8i/a xejivei, xal itpoc; op-ddc; auxrjv xe^ivei- 6p-&y] apa eaxlv fj utto ZEA- ndXiv, creel eu-deld tic, fj ZE eu-deldv xiva xrjv BA 8i/a xejivei, xal icpoc; op'da.z auxrjv xejivei- op^rj apa rj utto ZEB. e8e£)(i9r) 8e xal fj utto ZEA op'drj- iarj apa f) utto ZEA xfj utto ZEB fj eXdxxwv xrj (straight-line) CD, which is through the center and cuts in half the (straight-line) AB, which is not through the center, also cuts (AB) at right-angles. c D And so let CD cut AB at right-angles. I say that it also cuts (AB) in half. That is to say, that AF is equal to FB. For, with the same construction, since EA is equal to EB, angle EAF is also equal to EBF [Prop. 1.5]. And the right-angle AFE is also equal to the right-angle BFE. Thus, EAF and EFB are two triangles having two angles equal to two angles, and one side equal to one side — (namely), their common (side) EF, subtend- ing one of the equal angles. Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26]. Thus, AF (is) equal to FB. Thus, in a circle, if any straight-line through the cen- ter cuts in half any straight-line not through the center then it also cuts it at right-angles. And (conversely) if it cuts it at right-angles then it also cuts it in half. (Which is) the very thing it was required to show. Proposition 4 In a circle, if two straight-lines, which are not through the center, cut one another then they do not cut one an- other in half. Let ABCD be a circle, and within it, let two straight- lines, AC and BD, which are not through the center, cut one another at (point) E. I say that they do not cut one another in half. For, if possible, let them cut one another in half, such that AE is equal to EC, and BE to ED. And let the center of the circle ABCD have been found [Prop. 3.1], and let it be (at point) F, and let FE have been joined. Therefore, since some straight-line through the center, FE, cuts in half some straight-line not through the cen- ter, AC, it also cuts it at right-angles [Prop. 3.3]. Thus, FEA is a right-angle. Again, since some straight-line FE 73 ETOIXEIfiN y'. ELEMENTS BOOK 3 jjtei^ovi- oTtep eaxlv d§uvaxov. oux apa ai Ar, BA xe^ivouaiv 'Eav apa ev xuxXcp Suo eu-delai xe^ivwaiv dXXf|Xa<; ^f] 8la xou xevxpou ouaai, ou xe^ivouaiv dXXr|Xa<; Si/a - onep eSei BeT^ai. S . 'Eav Buo xuxXoi xe^ivcoaiv dXXr|Xouc, oux eaxai auxcov xo auxo xevxpov. Ado yap xuxXoi ol ABr, TAH xejivexcoaav dXXr|Xou<; xaxd xa B, T ar^ela. Xeyw, on oux eaxai auxGv xo auxo xevxpov. Ei ydp 8uvax6v, eaxw xo E, xal C7te£eu)(i!}G3 f\ EF, xal Bir^-dco f) EZH, ox exu/ev. xal eixei xo E a/jjielov xevxpov eaxl xou ABr xuxXou, Tar) eaxlv rj Er xfj EZ. rcaXiv, end xo E o/]jj.£lov xevxpov eaxl xou TAH xuxXou, I'ar) eaxlv f] Er xfj EH- eBeix^T) Be f] Er xal xfj EZ icny xal f) EZ apa xfj EH eaxiv Tar) f] eXdaaov xfj [leiCovi- onep eaxlv dBuvaxov. oux apa xo E ar)^eTov xevxpov eaxl xwv ABr, TAH xuxXcov. 'Eav apa Buo xuxXoi xeuvtoaiv dXXr]Xou<;, oux eaxiv cuts in half some straight-line BD, it also cuts it at right- angles [Prop. 3.3]. Thus, FEB (is) a right-angle. But FEA was also shown (to be) a right-angle. Thus, FEA (is) equal to FEB, the lesser to the greater. The very- thing is impossible. Thus, AC and BD do not cut one another in half. Thus, in a circle, if two straight-lines, which are not through the center, cut one another then they do not cut one another in half. (Which is) the very thing it was re- quired to show. Proposition 5 If two circles cut one another then they will not have the same center. For let the two circles ABC and CDC cut one another at points B and C. I say that they will not have the same center. For, if possible, let E be (the common center), and let EC have been joined, and let EFG have been drawn through (the two circles), at random. And since point E is the center of the circle ABC, EC is equal to EF. Again, since point E is the center of the circle CDG, EC is equal to EG. But EC was also shown (to be) equal to EF. Thus, EF is also equal to EG, the lesser to the greater. The very thing is impossible. Thus, point E is not 74 ETOIXEIfiN y'. ELEMENTS BOOK 3 auxwv to auTo xsvxpov onep eSsi SeT^ai. Auo yap xuxXoi ol ABr, TAE ecpanxso^woav dXXr]X«v xaxa to T CTTjjjieTov Xeyw, oti oux eaxai auTfiv to ai)TO xevTpov. El yap SuvaTov, eoto to Z, xal ETieCeu/'dw f] Zr, xal SirDcdw, £TU)(ev, f] ZEB. Tkel ouv to Z ar)[ieTov xevTpov eaxl tou ABr xuxXou, Xar] eaTiv r) Zr T/j ZB. TtdXw, ind to Z ar^ieTov xevTpov eoxl tou TAE xuxXou, for) saw f) Zr ttj ZE. £5d)cdr] 5e f] Zr T/j ZB Xor\- xal f) ZE dpa Tfj ZB eotiv Xai], f) eXAttcov Tfj jie^ovi- oTiep saw d5uvaTov. oux dpa to Z ar^eiov xevTpov sgtI twv ABr, TAE xuxXiov. 'Edv dpa 8uo xuxXoi ecpdnTCOVTai dXXr]X«v, oux ecrcai auTCSv to auTo xevTpov orcep sSei BsT^ai. 'Edv xuxXou era xfjt; 8ia^.£Tpou Xrjcp'df) ti ot][ieXo\i, 8 (ir| eoTi xevTpov tou xuxXou, arco 8e tou ar\\ie'iov izpbc, tov xuxXov TipocnuTTTwaiv su-delal Tiveg, jjlcyicttt) ^iev caTai, ecp' rjc; to xevTpov, eXa^iaT/) 8e f) Xomrj, tSv Be dXXov dtel f] £YY tov tffc §i a TO ° xevTpou Trj<; duwTspov piei^wv ecrav, 6uo 5e ^tovov Taai dno tou o/jjisiou TipoaneaouvTai 7tp6<; tov xuxXov £cp' exaTspa Trjg EXaxloTrjc;. the (common) center of the circles ABC and CDC Thus, if two circles cut one another then they will not have the same center. (Which is) the very thing it was required to show. Proposition 6 If two circles touch one another then they will not have the same center. For let the two circles ABC and CDE touch one an- other at point C. I say that they will not have the same center. For, if possible, let F be (the common center), and let FC have been joined, and let FEB have been drawn through (the two circles), at random. Therefore, since point F is the center of the circle ABC, FC is equal to FB. Again, since point F is the center of the circle CDE, FC is equal to FE. But FC was shown (to be) equal to FB. Thus, FE is also equal to FB, the lesser to the greater. The very thing is impos- sible. Thus, point F is not the (common) center of the circles ABC and CDE. Thus, if two circles touch one another then they will not have the same center. (Which is) the very thing it was required to show. Proposition 7 If some point, which is not the center of the circle, is taken on the diameter of a circle, and some straight- lines radiate from the point towards the (circumference of the) circle, then the greatest (straight-line) will be that on which the center (lies), and the least the remainder (of the same diameter) . And for the others, a (straight- line) nearer* to the (straight-line) through the center is always greater than a (straight-line) further away. And only two equal (straight-lines) will radiate from the point towards the (circumference of the) circle, (one) on each 75 ETOIXEIfiN y'. ELEMENTS BOOK 3 (side) of the least (straight-line). c bX \\ / z E V 1 'Eaxco xuxXoc 6 ABrA, 8id(iexpo<; Be auxou eaxw f\ A A, xal em xfjc A A eiXf|cp'dco ti a/jjieTov to Z, o \ir\ eaxi xevxpov xoO xuxXou, xevxpov 8e xou xuxXou eaxw xo E, xal duo xoO Z npo? xov ABrA xuxXov TtpoaTimxexcoaav eu'delai xivec od ZB, Zr, ZH- Xeyw, oxi ^teyiaxr) \ie\> eaxiv f\ ZA, eXa^iaxr) 8e f] ZA, xfiv 8e dXXwv f] ^tev ZB xfjc Zr ^tei^cov, f] 8e Zr xfjc ZH. , EKeCeu)fdwaav yap ai BE, TE, HE. xal eiiel Ttavxoc xpiywvou ai 8uo TtXeupal xfjc XoiTtfjc ^teiCovec eiaiv, ad dpa EB, EZ xfjc BZ jieiCovec eiaiv. Xor\ 8e f] AE xfj BE [ai dpa BE, EZ laai eiai xfj AZ]- ^lel^cov dpa f) AZ xfjc BZ. TtdXiv, enel Tar) eaxiv f) BE xfj TE, xoivf) 8e f] ZE, 8uo 6f] al BE, EZ Suoi xdic TE, EZ I'oai eiaiv. dXXd xal ycovia f\ uno BEZ yoviac xfjc Otto FEZ [ie[£«v pdaic dpa f] BZ pdaeioc xfjc rZ ^ei^tov eaxiv. 8id xd auxd 8rj xal f) TZ xfj? ZH ^et^wv eaxiv. ndXiv, enel ai HZ, ZE xfjc; EH [ici^ovec eiaiv, i'ar] 8e f] EH xfj EA, ai dpa HZ, ZE xfjc EA [iei^ovec eiaiv. xoivf) dcpflpriadtL) f] EZ- Xomf) dpa f) HZ Xomfjc xfjc; ZA [lei^cov eaxiv. (leyiax/) [lev dpa f) ZA, eXa)(iaxr) Se f) ZA, jiei^wv 6e f] txev ZB xfjc Zr, f) Be Zr xfjc ZH. Aeyio, oxi xal and xou Z arjjieiou 80o ^xovov laai Tipo- aiteaouvxai Tipoc xov ABrA xuxXov ecp' exdxepa xfjc ZA eXa/iaxrjg. auveaxdxw yap Ttpoc xrj EZ euiJeia xal xai Ttpoc auxfj a/j^ieitp xfi E xfj Otto HEZ yojvia I'ar) f) imb ZE9, xal eTie^eu)cdw r) Z6. enel ouv la/) eaxiv rj HE xfj E6, xoivf) 8e f) EZ, 8uo 8f) ai HE, EZ 8uai xdic 6E, EZ laai eiaiv xal ywvia f) Otto HEZ ywvia xfj Otto 0EZ lary pdaic dpa f] ZH pdaei xfj ZO Tar) eaxiv. Xeyo 5rj, oxi xfj ZH dXXr) for] ou KpoaneaeTxai TTpoc xov xuxXov duo xou Z ar^eiou. ei yap 8uvax6v, TrpoaTiiTixexo f] ZK. xal euel f\ ZK xfj ZH for] eaxiv, dXXd f] ZO xfj ZH [Tar] eaxiv], xal f] ZK dpa xfj ZO eaxiv I'ar], f] eyyiov xfjc 8id xou xevxpou xfj ducoxepov for)- oTiep dBuvaxov. oux dpa duo xou Z ar)[ieiou exepa xic Let ABCD be a circle, and let AZ) be its diameter, and let some point F, which is not the center of the circle, have been taken on AD. Let E be the center of the circle. And let some straight-lines, FB, FC, and FG, radiate from F towards (the circumference of) circle ABCD. I say that FA is the greatest (straight-line), FD the least, and of the others, FB (is) greater than FC, and FC than FG. For let BE, CE, and GE have been joined. And since for every triangle (any) two sides are greater than the remaining (side) [Prop. 1.20], EB and EF is thus greater than BF. And AE (is) equal to BE [thus, BE and EF is equal to AF]. Thus, AF (is) greater than BF. Again, since BE is equal to CE, and FE (is) common, the two (straight-lines) BE, EF are equal to the two (straight- lines) CE, EF (respectively). But, angle BEF (is) also greater than angle CEF} Thus, the base BF is greater than the base CF. Thus, the base BF is greater than the base CF [Prop. 1.24]. So, for the same (reasons), CF is also greater than FG. Again, since CF and FE are greater than EG [Prop. 1.20], and EG (is) equal to ED, GF and FE are thus greater than ED. Let EF have been taken from both. Thus, the remainder GF is greater than the re- mainder FD. Thus, FA (is) the greatest (straight-line), FD the least, and FB (is) greater than FC, and FC than FG. I also say that from point F only two equal (straight- lines) will radiate towards (the circumference of) circle ABCD, (one) on each (side) of the least (straight-line) FD. For let the (angle) FEH, equal to angle GEF, have been constructed on the straight-line EF, at the point E on it [Prop. 1.23], and let FH have been joined. There- fore, since GE is equal to EH, and EF (is) common, 76 ETOIXEIfiN y'. ELEMENTS BOOK 3 Tipoaueaaxai Tipoc; xov xuxXov iar) xfj HZ- (iia apa \i6vr\. 'Edv apa xuxXou era xfjc; Siauixpou X/jcp'dfj xi arjiMov, o (if] eaxi xevxpov xoO xuxXou, drab Be xou ar)[ieiou Ttpoc; xov xuxXov TtpoaraTixoaiv cu-dela! xivec;, (xeyiaxrj (xev eaxai, ecp' fjc; xo xevxpov, eXaxiaxr] 8s f) XoiTtf], xwv Se dXXwv del f) eyyiov xfjc; Sla xou xevxpou xfjc djtwxepov (iei£«v eaxiv, Suo Se (iovov I'aai duo xou auxou arpeiou TtpoaTteaoOvxai Ttpoc; xov xuxXov ecp' exdxepa xfjc eXa/iaxiqc;- ojtep eSei 8eTi;ai. the two (straight-lines) GE, EF are equal to the two (straight-lines) HE, EF (respectively). And angle GEF (is) equal to angle HEF. Thus, the base FG is equal to the base FH [Prop. 1.4]. So I say that another (straight- line) equal to FG will not radiate towards (the circumfer- ence of) the circle from point F. For, if possible, let FK (so) radiate. And since FK is equal to FG, but FH [is equal] to FG, FK is thus also equal to FH, the nearer to the (straight-line) through the center equal to the fur- ther away. The very thing (is) impossible. Thus, another (straight-line) equal to GF will not radiate from the point F towards (the circumference of) the circle. Thus, (there is) only one (such straight-line). Thus, if some point, which is not the center of the circle, is taken on the diameter of a circle, and some straight-lines radiate from the point towards the (circum- ference of the) circle, then the greatest (straight-line) will be that on which the center (lies), and the least the remainder (of the same diameter) . And for the oth- ers, a (straight-line) nearer to the (straight-line) through the center is always greater than a (straight-line) further away. And only two equal (straight-lines) will radiate from the same point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line). (Which is) the very thing it was required to show. t Presumably, in an angular sense. t This is not proved, except by reference to the figure. 'Edv xuxXou X/]cpiL>fj xi o/)[i.£lov exxoc;, duo Se xou a/][ie(ou 7tp6<; xov xuxXov Sia)(iL>Gaiv eu-delai xivec;, £>v [iia [lev 8id xou xevxpou, ai Se XoiTtai, &>c, exu)(ev, xGv (lev upog xf]v xoiX/jv mcpicpepeiav TtpoaraTixouacov eu'dsifiiv (leyiaxr] eaxiv f] Sid xou xevxpou, xwv Se aXXwv del f\ eyyiov xfjc; Sid xou xevxpou xfjc; draoxepov (ie(£«v eaxiv, xwv Se Ttpoc; xf]v xupxfjv nepicpepeiav upocnunxouafiv eui&eicov eXa^iaxr) [jlev eaxiv f) (iexac;u xou xe ar^efou xal xfjc; Sia^texpou, xwv Se aXXov del f] eyyiov t ^ eXayJaxrjc; xfjc; duwxepov eaxiv eXdxxcov, Suo Se fiovov laai and xou arpeiou TtpoaTteaoOvxai Ttpoc; xov xuxXov ecp' exdxepa xfjc; eXayJaxrjc;. "Eaxco xuxXoc; 6 ABr, xal xou ABr eiXfjcp'dw xi ay][ieTov exxoc; xo A, xal arc' auxou 8ir])fdwoav eu'delai xivec ai AA, AE, AZ, Ar, eaxw Se f] AA Sid xou xevxpou. Xeyto, oxi xwv [Lev Tipoc; xf]v AEZr xoiXrjv uepicpepeiav Ttpoara- Ttxouawv eu'deiwv (icyiaxr) [Lev eaxiv f) Sid xou xevxpou f) AA, pieiCwv Se f] [Lev AE xfjc; AZ f] Se AZ xfjc; Ar, xwv Se Tipoc xfjv 0AKH xupxfjv uepicpepeiav TtpoaTtiTtxouaCSv eu-deiCSv eXa)([axr) [Lev eaxiv r\ AH f) (lexacu xou ar^fjieiou xal xfjc; Siajiexpou xfjc AH, del Se f] eyyiov xfjc; AH eXa)(iaxir]C eXdxxwv eaxl xfjc; dicwxepov, f] [Lev AK xfjc; AA, f] Se AA Proposition 8 If some point is taken outside a circle, and some straight-lines are drawn from the point to the (circum- ference of the) circle, one of which (passes) through the center, the remainder (being) random, then for the straight-lines radiating towards the concave (part of the) circumference, the greatest is that (passing) through the center. For the others, a (straight-line) nearer^ to the (straight-line) through the center is always greater than one further away. For the straight-lines radiating towards the convex (part of the) circumference, the least is that between the point and the diameter. For the others, a (straight-line) nearer to the least (straight-line) is always less than one further away. And only two equal (straight- lines) will radiate from the point towards the (circum- ference of the) circle, (one) on each (side) of the least (straight-line). Let ABC be a circle, and let some point D have been taken outside ABC, and from it let some straight-lines, DA, DE, DF, and DC, have been drawn through (the circle), and let DA be through the center. I say that for the straight-lines radiating towards the concave (part of 77 ETOIXEIfiN y'. ELEMENTS BOOK 3 xfj? A9. A ElXf](p , dcL) yap to xevxpov xou ABr xuxXou xod eaxto to M- xai OTeCEUxtfwaav ai ME, MZ, Mr, MK, MA, M6. Kal ind Xai) eaxiv f] AM xfj EM, xoivf] Ttpoaxeia'dco f) MA- f) apa AA for) eaxi xai? EM, MA. dXX' ai EM, MA xfj? EA jieic^ove? daw xal f) AA apa xfj? EA ^eii^iov eaxiv. ndtXiv, end iar) eaxiv f] ME xfj MZ, xoivf] 8e f] MA, ai EM, MA apa xai? ZM, MA i'aai Eiaiv xai ytovia f] Otco EMA ywvia? xfj? utio ZMA ^.ei^cov eaxiv. pdai? dpa f] EA pdaew? xfj? ZA ^teii^ov eaxiv o^xoick Bf) 8eic;o^ev, oxi xai f] ZA xfj? EA ^iei^ov eaxiv ^eyiaxr] ^ev dpa f) AA, ^leii^ov 8e f) jiev AE xfjt; AZ, f] 8e AZ xfj? Ar. Kai etcei ai MK, KA xfj? MA jieiCove? eiaiv, iarj 8s r) MH xfj MK, Xomf] dpa f] KA Xonxrj? xfjt; HA jieiCwv eaxiv &axe f] HA xfj? KA eXdxxov eaxiv xai enei xpiywvou xou MAA em. (jiidt; xwv TtXeupGv xfjt; MA 8uo eu-deTai evxo? auveaxd , dr]aav ai MK, KA, ai dpa MK, KA xc5v MA, AA eXdxxove? eiaiv Iar] Se f] MK xfj MA- Xoircf] dpa f] AK Xomfj? xfj? AA eXdxxcov eaxiv. o^ioico? Bf] Sei^o^tev, oxi xai f] AA xfj? A8 eXdxxwv eaxiv eXa)(iaxr) [lev dpa f] AH, eXdxxwv Be f] [iev AK xfjt; AA f) 8e AA xfj? A9. Aey«, 6x1 xai 860 [lovov i'aai duo xou A ar)[ieiou npoaKeaouvxai n:p6? xov xuxXov ecp' exdxepa xfj? AH eXa)([axr]?- auveaxdxw rcpo? xfj MA eu'deia xai x£> npbc, auxfj ar)[ie[w iS M xfj Otto KMA ywvia i'ar) ywvia f) Otto AMB, xai CTieCeux'dw f) AB. xai etc el Xar] eaxiv f] MK xfj MB, xoivf] Be f] MA, 860 8f) ai KM, MA 860 xai? BM, MA the) circumference, AEFC, the greatest is the one (pass- ing) through the center, (namely) AD, and (that) DE (is) greater than DF, and DF than DC. For the straight-lines radiating towards the convex (part of the) circumference, HLKG, the least is the one between the point and the di- ameter AG, (namely) DG, and a (straight-line) nearer to the least (straight-line) DG is always less than one far- ther away, (so that) DK (is less) than DL, and DL than than DE. D For let the center of the circle have been found [Prop. 3.1], and let it be (at point) M [Prop. 3.1]. And let ME, MF, MC, MK, ML, and MH have been joined. And since AM is equal to EM, let MD have been added to both. Thus, AD is equal to EM and MD. But, EM and MD is greater than ED [Prop. 1.20]. Thus, AD is also greater than ED. Again, since ME is equal to MF, and MD (is) common, the (straight-lines) EM, MD are thus equal to FM, MD. And angle EMD is greater than angle FMD} Thus, the base ED is greater than the base FD [Prop. 1.24]. So, similarly, we can show that FD is also greater than CD. Thus, AD (is) the greatest (straight-line), and DE (is) greater than DF, and DF than DC. And since MK and KD is greater than MD [Prop. 1.20], and MG (is) equal to MK, the remainder KD is thus greater than the remainder GD. So GD is less than KD. And since in triangle MLD, the two inter- nal straight-lines MK and KD were constructed on one of the sides, MD, then MK and KD are thus less than ML and LD [Prop. 1.21]. And MK (is) equal to ML. Thus, the remainder DK is less than the remainder DL. So, similarly, we can show that DL is also less than DH. Thus, DG (is) the least (straight-line), and DK (is) less than DL, and £>£ than DiJ . I also say that only two equal (straight-lines) will radi- 78 ETOIXEIfiN y'. ELEMENTS BOOK 3 Taai eialv exaxepa exaxepqr xal ywvia f| uno KMA ycovia xfj UTio BMA iar)- pdaic; apa f) AK pdaei xfj AB for) eaxiv. Xeyw [8^] j oti xfj AK eu-deia aXXr] I'ar) ou TipoaTteaeTxai 7ip6<; xov xuxXov octio xou A ar^eiou. ei yap 8uvax6v, npo- aKiKxexo xal eaxw f] AN. end ouv f] AK xfj AN eaxiv Tar), dXX' f) AK xfj AB eaxiv Tar), xal f) AB apa xfj AN eaxiv Tar), f) eyyiov T ^ AH eXa)((axr](; xfj dn;«xepov [eaxiv] Tary orcep dSuvaxov e8e[)fdr]. oux apa rcXeiouc; fj Suo Taai 7ip6<; xov ABr xuxXov diio xoO A ar)[ieiou ecp' exdxepa xfj? AH eXa)((oxr]<; TtpoaTteaouvxai. 'Eav apa xuxXou Xrjcp'dfj xi ar)(ieTov exxoc, arco 5e xou ay][ie'iou npbc, xov xuxXov 8ia)fd«aiv eu'delai xivec;, Sv ^ua ^.sv 8id xou xevxpou ai 8e Xomai, (be; exu)(£v, xcov ^iev 7ip6<; xf]v xo(Xr)v uepicpepeiav Tipoarauxouawv sutJeiGv (jieyiaxr] [lev eaxiv f) 8ia xou xevxou, xfiv 8e aXXwv del f] eyyiov xfjc; Sid xou xevxpou xfj<; djiwxepov [ie(C«v eaxiv, xwv 8e 7tp6<; xf]v xupxf]v uepicpepeiav KpoaTiiTixouawv euiJeiGv eXa)([axr) ^ev eaxiv f) ^texa^u xou xe ar^eiou xal xfj? 8ia^texpou, iSv 8e dXXwv del f] eyyiov xfjc; eXaxiaxrjc xfjc; dftoxepov eaxiv eXdxxMv, 8uo 8e ^.ovov Taai duo xou arj^ieiou Tipoaueaouvxai 7ip6<; xov xuxXov ecp' exdxepa xfjc eXaxiax/]<;- orcep e8ei Bel^ai. ate from point D towards (the circumference of) the cir- cle, (one) on each (side) on the least (straight-line), DG. Let the angle DMB, equal to angle KMD, have been constructed on the straight-line MD, at the point M on it [Prop. 1.23], and let DB have been joined. And since MK is equal to MB, and MD (is) common, the two (straight-lines) KM, MD are equal to the two (straight- lines) BM, MD, respectively. And angle KMD (is) equal to angle BMD. Thus, the base DK is equal to the base DB [Prop. 1.4]. [So] I say that another (straight- line) equal to DK will not radiate towards the (circum- ference of the) circle from point D. For, if possible, let (such a straight-line) radiate, and let it be DN. There- fore, since DK is equal to DN, but DK is equal to DB, then DB is thus also equal to DN, (so that) a (straight- line) nearer to the least (straight-line) DG [is] equal to one further away. The very thing was shown (to be) im- possible. Thus, not more than two equal (straight-lines) will radiate towards (the circumference of) circle ABC from point D, (one) on each side of the least (straight- line) DG. Thus, if some point is taken outside a circle, and some straight-lines are drawn from the point to the (circumfer- ence of the) circle, one of which (passes) through the cen- ter, the remainder (being) random, then for the straight- lines radiating towards the concave (part of the) circum- ference, the greatest is that (passing) through the center. For the others, a (straight-line) nearer to the (straight- line) through the center is always greater than one fur- ther away. For the straight-lines radiating towards the convex (part of the) circumference, the least is that be- tween the point and the diameter. For the others, a (straight-line) nearer to the least (straight-line) is always less than one further away. And only two equal (straight- lines) will radiate from the point towards the (circum- ference of the) circle, (one) on each (side) of the least (straight-line). (Which is) the very thing it was required to show. t Presumably, in an angular sense. | This is not proved, except by reference to the figure. ft'. 'Eav xuxXou Xrjcp'dfj xi ar\\ie1ov evxoc;, olko 8e xou arj^eiou 7ipo<; xov xuxXov Ttpoaiuitxwai nkeiouq fj Suo Taai eu-delai, xo Xrj^ev arpeTov xevxpov eaxl xou xuxXou. 'Eaxco xuxXoc; 6 ABr, evxoc 8e auxou ar)[ieTov xo A, xal dTio xou A 7ip6<; xov ABr xuxXov upoaraTixexcoaav TtXeiouc; f] 8uo Taai eu-delai al AA, AB, AE Xey«, oxi xo A ar)[ieTov xevxpov eaxl xou ABr xuxXou. Proposition 9 If some point is taken inside a circle, and more than two equal straight-lines radiate from the point towards the (circumference of the) circle, then the point taken is the center of the circle. Let ABC be a circle, and D a point inside it, and let more than two equal straight-lines, DA, DB, and DC, ra- diate from D towards (the circumference of) circle ABC. 79 ETOIXEIfiN y'. ELEMENTS BOOK 3 A O TkeCetijcdcooav yap ai AB, Br xal TeTuVjcydtoaav Bi^a xaxd xa E, Z ar^eTa, xal ETuCeux'deiaai al EA, ZA 8if])(Tf)waav era xa H, K, 0, A ar^ela. Tkei ouv igt] eaxiv #] AE xrj EB, xoivr) 8e #] EA, 8uo 8f) al AE, EA 660 xau; BE, EA laai eialv xal pdau; f] AA pdasi xfj AB tar) - y«v(a dpa f) utto AEA ycovia xfj Otto BEA iar] eaxlv op'dr] dpa exaxepa xwv O716 AEA, BEA yoviwv f) HK dpa xrjv AB xe^ivei Sixa xal npbc, op^dc. xal ejxel, eav ev xuxXco EuiDeTd xu; euiDeldv xiva 8ixa xe xal 7tp6<; 6piDd<; xe^ivr], £7il xfj<; xe[ivo6ar)<; eaxl xo xevxpov xou xuxXou, era xfj? HK dpa eaxl xo xevxpov xou xuxXou. Sid xd auxd 8r) xal zid xrjc 0A eaxi xo xevxpov xou ABr xuxXou. xal ouSsv exepov xoivov exouaiv ai HK, OA eu'de'iai f\ xo A arpeTov xo A dpa ar^elov xevxpov eaxl xou ABr xuxXou. 'Eav dpa xuxXou XricpiDfj xi ar^elov evxoc, duo 8e xou ar\\ieiov izpbz xov xuxXov Ttpoarararwai uXeiouc; rj 860 laai eu-delai, xo Xr)cp{>ev ar^elov xevxpov eaxl xou xuxXou- oTtep e8ei BeT^ai. I say that point D is the center of circle ABC. B/ F __\ k — A^ D H For let AB and BC have been joined, and (then) have been cut in half at points E and F (respectively) [Prop. 1.10]. And ED and FD being joined, let them have been drawn through to points G, K, H, and L. Therefore, since AE is equal to EB, and ED (is) com- mon, the two (straight-lines) AE, ED are equal to the two (straight-lines) BE, ED (respectively). And the base DA (is) equal to the base DB. Thus, angle AED is equal to angle BED [Prop. 1.8]. Thus, angles AED and BED (are) each right-angles [Def. 1.10]. Thus, GK cuts AB in half, and at right-angles. And since, if some straight-line in a circle cuts some (other) straight-line in half, and at right-angles, then the center of the circle is on the former (straight-line) [Prop. 3.1 corr.], the center of the circle is thus on GK. So, for the same (reasons), the center of circle ABC is also on HL. And the straight-lines GK and HL have no common (point) other than point D. Thus, point D is the center of circle ABC. Thus, if some point is taken inside a circle, and more than two equal straight-lines radiate from the point to- wards the (circumference of the) circle, then the point taken is the center of the circle. (Which is) the very thing it was required to show. l . KuxXoc; xuxXov ou xejivei xaxd nXebva orj^iela fj 860. EE yap 8uvax6v, xuxXog 6 ABT xuxXov xov AEZ xe^tvexM xaxd nXeiova ar)[ieTa fj 860 xa B, H, Z, O, xal eTuCeuyjdeTaai ai B9, BH Sfya xe^ivecrdwaav xaxd xd K, A ar)[ieTa- xal aito xwv K, A xau; B9, BH Ttp6<; opMc; dx^elaai ai KT, AM 8ir])cdwoav era xd A, E ar)^eTa. Proposition 10 A circle does not cut a(nother) circle at more than two points. For, if possible, let the circle ABC cut the circle DEF at more than two points, B, G, F, and H . And BH and BG being joined, let them (then) have been cut in half at points K and L (respectively). And KC and LM be- ing drawn at right-angles to BH and BG from K and L (respectively) [Prop. 1.11], let them (then) have been drawn through to points A and E (respectively). 80 ETOIXEIfiN y'. ELEMENTS BOOK 3 Tkel ouv ev xuxXw xfi ABr eui)eTd tic, f) AT su'dei&v xiva xf]v B9 h'v/O- xal Ttpoc; op-ddc; xe^jivei, era xfjc; Ar apa eaxl to xevxpov tou ABr xuxXou. TtdXiv, itnel ev xuxXcp xG auxG tu ABr £015616; tic; f] NS eu'deldv xiva xf)v BH 8[)(a xal Ttpoc; op'dac; xe^vei, era xfjc; NS apa eaxl xo xevxpov xoO ABr xuxXou. eBeix'dr) 8e xal era xfjc; Ar, xal xax' oi)8ev au^pdXXouaiv ai Ar, NS eu-delai fj xaxd xo O xo O dpa at][ieTov xevxpov laxl xou ABr xuxXou. o^toiwc; Sf] 8eic;ouev, oxi xal xou AEZ xuxXou xevxpov eaxl xo O 8uo apa xuxXwv Teuv6vT«v dXXfjXouc; tGv ABr, AEZ xo auxo eaxi xevxpov xo O ojtep eaxlv d56vaxov. Oux apa xuxXoc; xuxXov xe^ivei xaxd nXetova arj(ieTa f\ 8uo- onep e5ei Belial. N L \o Therefore, since in circle ABC some straight-line AC cuts some (other) straight-line BH in half, and at right-angles, the center of circle ABC is thus on AC [Prop. 3.1 corr.]. Again, since in the same circle ABC some straight-line NO cuts some (other straight-line) BG in half, and at right-angles, the center of circle ABC is thus on NO [Prop. 3.1 corr.]. And it was also shown (to be) on AC. And the straight-lines AC and NO meet at no other (point) than P. Thus, point P is the center of circle ABC. So, similarly, we can show that P is also the center of circle DEF. Thus, two circles cutting one an- other, ABC and DEF, have the same center P. The very- thing is impossible [Prop. 3.5]. Thus, a circle does not cut a(nother) circle at more than two points. (Which is) the very thing it was required to show. lot'. 'Edv 8uo xuxXoi ecpdnxwvxai dXXf]X«v evToc;, xal Xr)tfdfj auxwv xd xevxpa, f) era xd xevxpa auxwv eraCeuyvu^ievr) eMela xal exPaXXo^tevr] era xrjv ouvacprjv TieoeTxai tGv xuxXgjv. Auo yap xuxXoi oi ABT, AAE ecpaTixeo-dwaav dXXf|Xwv evxo<; xaxd xo A ar^elov, xal elX^cp-dco xou y.kv ABT xuxXou xevxpov xo Z, xou 8e AAE xo H- Xeyco, oxi f\ duo xou H era xo Z era^euYvu^evr) eu^ela ex[3aXXo^evr) era xo A neaelxai. Mr] yap, dXX' et Suvaxov, ranxexw Gc; f] ZH9, xal ejie^eu)cd«aav ai AZ, AH. 'Eracl ouv ai AH, HZ xfjc; ZA, xouxeaxi xfjc; Z9, (leiCovec; eioiv, xoivr) dcprpfjadio f] ZH- XoiTtf] apa f] AH Xomfjc; xfjc H0 [is^cov eaxiv. for) 8e f] AH xfj HA- xal f] HA apa xfjc; H8 ^teiCwv eaxlv f] eXdxxwv xfjc; [leiCovoc;- oracp eaxlv dSuvaxov oux apa f) drab xou Z era xo H eraCeuyvu^ievr) eu-dela exxoc; ueaelxai- xaxd xo A apa era xfjc; auvacpfjc racaelxai. Proposition 11 If two circles touch one another internally, and their centers are found, then the straight-line joining their cen- ters, being produced, will fall upon the point of union of the circles. For let two circles, ABC and ADE, touch one another internally at point A, and let the center F of circle ABC have been found [Prop. 3.1], and (the center) G of (cir- cle) ADE [Prop. 3.1]. I say that the straight-line joining G to F, being produced, will fall on A. For (if) not then, if possible, let it fall like FGH (in the figure), and let AF and AG have been joined. Therefore, since AG and GF is greater than FA, that is to say FH [Prop. 1.20], let FG have been taken from both. Thus, the remainder AG is greater than the re- mainder GH. And AG (is) equal to GD. Thus, GD is also greater than GH, the lesser than the greater. The very thing is impossible. Thus, the straight-line joining F to G will not fall outside (one circle but inside the other) . Thus, it will fall upon the point of union (of the circles) 81 ETOIXEIfiN y'. ELEMENTS BOOK 3 'Edv dpa 860 xuxXoi ecpditxcovxai dXXf|XGJv evxoc;, [xal Xrjcydrj auxfiv xd xevxpa], f] Ira xd xevxpa auxwv era^eu- yvupievr] euiiteTa [xal expaXXojievr)] era xf]v auvacpfjv rceaelxai twv xuxXtov ouep e8ei BeT^ai. LP'. 'Eav 860 xuxXoi ecpdnxtovxai dXXf|Xt>)v exxoc;, f] era xa xevxpa auxov era^euyvujievT] 6id xfjc eixacpfjc; eXeuaexai. B E Auo yap xuxXoi oi ABr, AAE ecpaTtxea-dtoaav dXXf|XtL>v exxoc; xaxd xo A arjjielov, xal elXf]cpi9M xoO [Lev ABT xevxpov xo Z, xou 8e AAE xo H' Xeyw, oxi f) drab xoO Z era xo H era^euyvunevr) eu-dela 8id xfjt; xaxd xo A eixacpfjc; eXeuaexai. Mf] yap, dXX'' el 8uvax6v, epxecrdw «<; f) ZrAH, xal CTieCeux'Swaav ai AZ, AH. Tkel ouv xo Z ar)[ieTov xevxpov eaxl xou ABr xuxXou, Xai] eaxlv f] ZA xfj Zr. raxXiv, enel xo H ar)[ieTov xevxpov eaxl xou AAE xuxXou, iar\ eaxlv f) HA xfj HA. eSeix^ at point A. H Thus, if two circles touch one another internally, [and their centers are found], then the straight-line joining their centers, [being produced], will fall upon the point of union of the circles. (Which is) the very thing it was required to show. Proposition 12 If two circles touch one another externally then the (straight-line) joining their centers will go through the point of union. B For let two circles, ABC and ADE, touch one an- other externally at point A, and let the center F of ABC have been found [Prop. 3.1], and (the center) G of ADE [Prop. 3.1] . I say that the straight-line joining F to G will go through the point of union at A. For (if) not then, if possible, let it go like FCDG (in the figure), and let AF and AG have been joined. Therefore, since point F is the center of circle ABC, FA is equal to FC. Again, since point G is the center of circle ADE, GA is equal to GD. And FA was also shown 82 ETOIXEIfiN y'. ELEMENTS BOOK 3 Be xal f] ZA xrj ZT lory ai apa ZA, AH xdu; Zr, HA 1'aai eto£v woxe okr\ f) ZH xwv ZA, AH ja.e[£cov eaxiv dXXa xal eXdxxcov onep eaxlv dSuvaxov. oux apa f] arco xou Z era xo H era^euyvujjievr] eGi&e'ia 8id xrjg xaxd xo A eraxcpfjc; oux eXeuoexai- Si' aGxrjc; apa. 'Edv apa 860 xuxXoi ecpdnxiovxai dXXrjXwv exxoc;, f) era xd xevxpa auxGv eraCeuyvu^evr] [eO'dela] 81a xrjc; enacprjc eXeuaexai - ouep eSei 8eTc;ai. EE yap 8uvax6v, xuxXoc; 6 ABrA xuxXou xou EBZA ecpanxecrdco upoxepov evxoc; xaxd nXeiova ar)|jieTa fj ev xd A, B. Kal EiX^cpdco xou ^iev ABrA xuxXou xevxpov xo H, xou 8e EBZA xo 6. ; H apa duo xou H era xo © eTu^euyvujievr] era xd B, A TteaeTxai. tuttxcxm (be; #j BH6A. xal end xo H a/jjietov xevxpov eoxl xou ABrA xuxXou, Tor) eaxlv #j BH xfj HA- (lei^tdv apa f) BH xrjc; OA- ttoXXG apa (jiei^wv f] B9 xrjc; OA. ndXiv, knzi xo 6 a/jjielov xevxpov eaxl xou EBZA xuxXou, Tar) eaxlv f) B6 xfj OA- eBeixii/] Be auxfjg xal ttoXXG jiei^cov ojiep douvaxov oux apa xuxXog xuxXou ecpdnxexai evxoc xaxd TtXeiova a/jjiela fj ev. Aeyco orj, oxi ou6e exxoc;. Ei yap Suvaxov, xuxXoc; 6 ArK xuxXou xou ABrA ecpaTixeaiJco exxoc; xaxd TtXeiova ar)(iela f] ev xd A, T, xal eTie^eu)cdw f\ AT. "Enzl ouv xuxXgjv x£>v ABrA, ArK ei'XrjTtxai km. xrjc; Ttepicpepeiac; exaxepou 860 xu)(6vxa or)(ieTa xd A, T, f) era xd ar)U.eTa era^euyvu^evr) cu^ela evxoc; exaxepou neaelxar dXXa xou [iev ABrA evxoc; eneaev, xou Be ArK exxoc;- ojiep axoTtov oux apa xuxXoc; xuxXou ecpdnxexai exxoc; xaxd TiXebva ar^ela fj ev. eBeiy^r] Be, 6x1 ouBe evxoc;. (to be) equal to FC. Thus, the (straight-lines) FA and AG are equal to the (straight-lines) FC and GD. So the whole of FG is greater than FA and AG. But, (it is) also less [Prop. 1.20]. The very thing is impossible. Thus, the straight-line joining F to G cannot not go through the point of union at A. Thus, (it will go) through it. Thus, if two circles touch one another externally then the [straight-line] joining their centers will go through the point of union. (Which is) the very thing it was re- quired to show. Proposition 13 A circle does not touch a(nother) circle at more than one point, whether they touch internally or externally For, if possible, let circle ABDC 1 * touch circle EBFD — first of all, internally — at more than one point, D and B. And let the center G of circle ABDC have been found [Prop. 3.1], and (the center) H of EBFD [Prop. 3.1]. Thus, the (straight-line) joining G and H will fall on B and D [Prop. 3.11]. Let it fall like BGHD (in the figure) . And since point G is the center of circle ABDC, BG is equal to GD. Thus, BG (is) greater than HD. Thus, BH (is) much greater than HD. Again, since point H is the center of circle EBFD, BH is equal to HD. But it was also shown (to be) much greater than it. The very thing (is) impossible. Thus, a circle does not touch a(nother) circle internally at more than one point. So, I say that neither (does it touch) externally (at more than one point) . For, if possible, let circle ACK touch circle ABDC externally at more than one point, A and C. And let AC have been joined. Therefore, since two points, A and C, have been taken at random on the circumference of each of the circles ABDC and ACK , the straight-line joining the points will fall inside each (circle) [Prop. 3.2]. But, it fell inside ABDC, and outside ACK [Def. 3.3]. The very thing 83 ETOIXEIfiN y'. ELEMENTS BOOK 3 KuxXog dpa xuxXou oux ecpditxexai xaxd TtXeiova arj^ieTa rj [xotd'] ev, edv xe evxoc edv xe exxoc; ecpdnxrjxai' ojtep e8ei 8eTc;ai. t The Greek text has "ABCD", which is obviously a mistake. 18'. 'Ev xuxXcp al faai euifteTai faov djie/ouaiv and xou xevxpou, xai al faov dne/ouaai dno xou xevxpou faai dXXrjXaic eiaiv. A "Eaxw xuxXoc; 6 ABrA, xai ev auxco faai eO'delai eaxco- aav al AB, EA- Xeyco, oxi al AB, EA faov aTte)(ouaiv arco xou xevxpou. EiXr)cp , dco yap xo xevxov xou ABrA xuxXou xdi eaxw xo E, xai and xou E era xd<; AB, TA xdiJexoi fjx'dwcjav al EZ, EH, xdi eirsCeOx'OwCTav al AE, EE 'Eitei ouv eicdeTd xu; 8la xou xevxpou f] EZ euiDeTdv xiva ^.f] 8ia xou xevxpou xf)v AB Ttpoc; op'da? xe^ivei, xai 8[)(a auxfjv xe^vei. far] apa f] AZ xfj ZB- SixcXfj apa f] AB xrj<; AZ. 8id xd auxd Sf] xai f] TA xrj<; TH eaxi SiirArj" xai eaxiv far] f) AB xfj EA- far) apa xai f) AZ xfj TH. xai ckci far] eaxiv f] AE xfj EE, iaov xai xo d^o xfj<; AE xw duo xrj<; EE dXXd tu [i£v arco xrj<; AE laa xd duo xwv AZ, EZ' op'dr) yap f) 7ip6<; xw Z ywvia- xw 8e diio xfj? Er laa xd diio xwv EH, HE op'df] yap f) 7tp6<; iu H y«v(a- xd apa dfto xtov AZ, ZE faa eaxi xoTc duo xwv TH, HE, Sv xo duo xrj<; AZ iaov eaxi tu drco xfj? TH- far) yap eaxiv f] AZ xfj ITE Xoikov apa xo duo xfjc; ZE tw duo xrjc EH faov eaxiv far] apa f] EZ xfj EH. ev Se xuxXw iaov dTte^eiv duo xou xevxpou eu-delai Xeyovxai, oxav al and xou xevxpou in auxdc; xdiJexoi dyo^tevai faai CSaiv al apa AB, EA faov duexouaiv ano xou xevxpou. AXXd 8f) al AB, EA eu'delai faov dnexexwaav duo xou xevxpou, xouxeaxiv far) eaxco f) EZ xfj EH. Xeyw, oxi far] eaxi xai f] AB xfj EA. (is) absurd. Thus, a circle does not touch a(nother) circle externally at more than one point. And it was shown that neither (does it) internally. Thus, a circle does not touch a(nother) circle at more than one point, whether they touch internally or exter- nally. (Which is) the very thing it was required to show. Proposition 14 In a circle, equal straight-lines are equally far from the center, and (straight-lines) which are equally far from the center are equal to one another. B/ \ \ E^G \ F \ \\ / A Let ABD0 be a circle, and let AB and CD be equal straight-lines within it. I say that AB and CD are equally far from the center. For let the center of circle ABDC have been found [Prop. 3.1], and let it be (at) E. And let EF and EG have been drawn from (point) E, perpendicular to AB and CD (respectively) [Prop. 1.12]. And let AE and EC have been joined. Therefore, since some straight-line, EF, through the center (of the circle), cuts some (other) straight-line, AB, not through the center, at right-angles, it also cuts it in half [Prop. 3.3]. Thus, AF (is) equal to FB. Thus, AB (is) double AF. So, for the same (reasons), CD is also double CG. And AB is equal to CD. Thus, AF (is) also equal to CG. And since AE is equal to EC, the (square) on AE (is) also equal to the (square) on EC. But, the (sum of the squares) on AF and EF (is) equal to the (square) on AE. For the angle at F (is) a right- angle [Prop. 1.47]. And the (sum of the squares) on EG and GC (is) equal to the (square) on EC. For the angle at G (is) a right-angle [Prop. 1.47]. Thus, the (sum of the squares) on AF and FE is equal to the (sum of the squares) on CG and GE, of which the (square) on AF is equal to the (square) on CG. For AF is equal to CG. 84 ETOIXEIfiN y'. ELEMENTS BOOK 3 Toiv ydp auxfiv xaxaoxeuaaiDevTWv 6[io[cl><; 8eic;o^.ev, oxi SiTtXfj eaxiv f) jiev AB tt^c: AZ, f) 8e TA xfjc; TH- xal end Tar) eaxiv f) AE xfj TE, i'aov eaxl xo ano xfjc; AE x« arco xfjc; TE' dXXd xco [iev and xfjc; AE Taa eaxl xd duo xc5v EZ, ZA, xcp Se (xtco xfjc; TE Taa xd duo xfiv EH, HT. xd dpa arco xfiv EZ, ZA foot eaxl xolc; dno x£5v EH, HT- Sv xo arco xfjc; EZ xfi duo xfjc; EH eaxiv iaov Tar) yap /) EZ xfj EH' Xoitiov dpa xo duo xfjc; AZ i'aov eaxl xw duo xfjc; TH- lar\ dpa f) AZ xfj TH- xal eaxi xfjc; jiev AZ SiTtXfj f] AB, xfjc; Se TH SiTtXfj f) TA- Tar) dpa f] AB xfj TA. 'Ev xuxXw dpa al Taai eu-delai I'aov aTtexouaiv duo xou xevxpou, xal ai I'oov dnexouaai duo xou xevxpou Taai dXXfjXaic; eto£v OTtep e8ei Bel^ai. t The Greek text has "ABCD", which is obviously a mistake. IS . 'Ev xuxXcp (ieyiaxT] ^ev f] Bidjiexpoc;, xwv 8e dXXcov del f] eyyiov xou xevxpou xfjc; duwxepov ^.si^wv saxiv. 'Eaxco xuxXoc; 6 ABrA, 8id[iexpoc; 8e auxou eox« #j A A, xevxpov 8e xo E, xal eyyiov ^xev xfjc; AA 8ia^texpou eaxo f) Br, aTioxepov 8e f\ ZH- Xeyw, oxi [leytaxrj [Lev eoxiv f] AA, ^ielC«v 8e f] Br xfjc; ZH. "H)edtL>aav yap dito xou E xevxpou era xdc; Br, ZH xdiftexoi ai E6, EK. xal ensl eyyiov ^tev xou xevxpou eaxiv f] Br, dnwxepov 8e f] ZH, ^ei^wv dpa f] EK xfjc; E6. xeirj'do xfj E9 lay] f] EA, xal 8id xou A xfj EK Tipoc; op-Quc, d)fdeTaa f] AM 8if]X'S« era to N, xal eTte^eux^waav al ME, EN, ZE, EH. Kal CTtel iat] eaxiv f] E9 xfj EA, lot] eaxl xal f] BT xfj MN. TtdXiv, euel iarj eaxiv f] ^xev AE xfj EM, f] 8e EA x^ EN, f) dpa AA xalc; ME, EN I'ar) eaxiv. dXX'' ai jiev ME, EN xfjc; MN ^.eiCovec; eiaiv [xal f) AA xfjc; MN (iciCwv eaxiv], iar] 8e f) MN xfj Br- f) AA dpa xfjc; Br [ie(Cwv eaxiv. xal CTtel 8uo ai ME, EN 8uo xalc; ZE, EH i'aai eiaiv, xal ywvla f] bub MEN ywviac; xfjc; UTto ZEH ^lei^wv [eaxiv], [3daic; dpa Thus, the remaining (square) on FE is equal to the (re- maining square) on EG. Thus, EF (is) equal to EG. And straight-lines in a circle are said to be equally far from the center when perpendicular (straight-lines) which are drawn to them from the center are equal [Def. 3.4]. Thus, AB and CD are equally far from the center. So, let the straight-lines AB and CD be equally far from the center. That is to say, let EF be equal to EG. I say that AB is also equal to CD. For, with the same construction, we can, similarly, show that AB is double AF, and CD (double) CG. And since AE is equal to CE, the (square) on AE is equal to the (square) on CE. But, the (sum of the squares) on EF and FA is equal to the (square) on AE [Prop. 1.47]. And the (sum of the squares) on EG and GC (is) equal to the (square) on CE [Prop. 1.47]. Thus, the (sum of the squares) on EF and FA is equal to the (sum of the squares) on EG and GC, of which the (square) on EF is equal to the (square) on EG. For EF (is) equal to EG. Thus, the remaining (square) on AF is equal to the (re- maining square) on CG. Thus, AF (is) equal to CG. And AB is double AF, and CD double CG. Thus, AB (is) equal to CD. Thus, in a circle, equal straight-lines are equally far from the center, and (straight-lines) which are equally far from the center are equal to one another. (Which is) the very thing it was required to show. Proposition 15 In a circle, a diameter (is) the greatest (straight-line), and for the others, a (straight-line) nearer to the center is always greater than one further away. Let ABCD be a circle, and let AD be its diameter, and E (its) center. And let BC be nearer to the diameter AD ,t and FG further away. I say that AD is the greatest (straight-line), and BC (is) greater than FG. For let EH and EK have been drawn from the cen- ter E, at right-angles to BC and FG (respectively) [Prop. 1.12]. And since BC is nearer to the center, and FG further away, EK (is) thus greater than EH [Def. 3.5]. Let EL be made equal to EH [Prop. 1.3]. And LM being drawn through L, at right-angles to EK [Prop. 1.11], let it have been drawn through to N. And let ME, EN, FE, and EG have been joined. And since EH is equal to EL, BC is also equal to MN [Prop. 3.14]. Again, since AE is equal to EM, and ED to EN, AD is thus equal to ME and EN. But, ME and EN is greater than MN [Prop. 1.20] [also AD is 85 ETOIXEIfiN y'. ELEMENTS BOOK 3 f) MN pdaecoc; xfjc ZH ^ei^tov eaxlv. dXXd fj MN xfj Br greater than MJV], and MJV (is) equal to SC. Thus, AL> sSeixdr] Tar) [xod fj Br xfjc ZH ^ie[£«v eaxlv] . (jlsylcttt) [iev is greater than BC. And since the two (straight-lines) dpa fj AA Bid^iexpoc;, liel^tov 8s fj Br xfjc; ZH. ME, EN are equal to the two (straight-lines) EE, EG (respectively), and angle MEN [is] greater than angle FEG, X the base MN is thus greater than the base FG [Prop. 1.24]. But, MN was shown (to be) equal to BC [(so) BC is also greater than FG]. Thus, the diameter AD (is) the greatest (straight-line), and BC (is) greater than FG. 'Ev xuxXw dpa ^syiaxr) jiev eaxiv f| 8id[iexpoc;, xfiv Be: Thus, in a circle, a diameter (is) the greatest (straight- dXXwv del f] syyiov xoO xevxpou xfjc ditwxepov (jeiCuv eaxlv line), and for the others, a (straight-line) nearer to the oTtep e8ei Bei^ai. center is always greater than one further away. (Which is) the very thing it was required to show. f Euclid should have said "to the center", rather than "to the diameter AD", since BC, AD and FG are not necessarily parallel, t This is not proved, except by reference to the figure. H xfj Sia^expw xou xuxXou 7ip6<; opMc; dm' dxpac; dyo^ievr] exxoc; TteaeTxai xou xuxXou, xal sic; xov [iexac;u xoitov xfjc; xe eu'delac; xal xfjc; Ttepicpspeiae; exepa eu-dela ou TtapejjmeaeTxai, xal fj [lev xou fjuixuxXfou Y wv i a aTtdarjc; Y«v(a<; 6e;elac; eu-duYpd^ou ^teiCtov eaxlv, f) 8e XoiTtf) eXdxxwv. TCaxo xuxXoc; 6 ABT nepl xevxpov xo A xal 8id^texpov xfjv AB- Xsy", 8xi fj duo xou A xfj AB Ttpoc; opMc; an' dxpac; dYo^ievr] exxoc; neaeTxai xou xuxXou. Mf] Ydp, dXX' el 8uvax6v, mnisxw evxoc; wc; f) TA, xal £Tie^£U)cd« fj AT. 'EticI larj eaxlv f] AA xfj AT, ia/j eaxl xal ywvla fj uti:6 AAr ywvia xfj utto ArA. opiSfj 8s fj utto AAr- opiSfj dpa xal fj Otto ArA' xpiytiivou 8f) xou ArA al Buo y<^io(1 oti utto AAT, ATA 6uo op-dalc; I'oai eialv oTiep eaxlv dSuvaxov. oux dpa f) arco xou A at]y.e'\.ou xfj BA Ttpoc; 6p$a<; aYo^evr) evxoc; neaeTxai xou xuxXou. by.oioic, 8fj SeTd;o^ev, oxi ou8' stxl xfjc; nspicpspeiac;- exxoc; dpa. Proposition 16 A (straight-line) drawn at right-angles to the diameter of a circle, from its end, will fall outside the circle. And another straight-line cannot be inserted into the space be- tween the (aforementioned) straight-line and the circum- ference. And the angle of the semi-circle is greater than any acute rectilinear angle whatsoever, and the remain- ing (angle is) less (than any acute rectilinear angle) . Let ABC be a circle around the center D and the di- ameter AB. I say that the (straight-line) drawn from A, at right-angles to AB [Prop 1.11], from its end, will fall outside the circle. For (if) not then, if possible, let it fall inside, like CA (in the figure), and let DC have been joined. Since DA is equal to DC, angle DAG is also equal to angle ACD [Prop. 1.5]. And DAG (is) a right-angle. Thus, ACD (is) also a right-angle. So, in triangle ACD, the two angles DAG and ACD are equal to two right- angles. The very thing is impossible [Prop. 1.17]. Thus, the (straight-line) drawn from point A, at right-angles 8G ETOIXEIfiN y'. ELEMENTS BOOK 3 B IIiTcxexco d>c f] AE- Xeyio 8r], oxi eic xov jiexacu xotiov xfjc xe AE eu'deiac xal xfjc T9A Txeptcpepeta^ exepa eu-dela ou Tiape^iTceaeTxai. Et yap 8uvax6v, Tiape|iTiiTixex6) 6? f] ZA, xal fjx'dco &ti6 xoO A ar](!£iou era. xfjv ZA xdcdsxcx; f) AH. xal CTiel op-df] eaxiv f) utio AHA, eXdxxtov 8e op'dfjc f] utio AAH, ^teii^v apa f) A A xfjc AH. iar) Be f] A A xfj A6- (j.£l^cov apa f] A0 xfjc AH, f] eXdxxwv xfjc ^el^ovoc oTiep eaxlv dBuvaxov. oux apa etc xov liexacu xotcov xfjc xe eMeiac xal xfjc Tiepicpepelac exepa eu-dela Tiape^iTceaeTxai. Aeyco, oxi xal f) \±ev xoO fjuixuxXlou ycovia f] Tiepiexo^ievr] utio xe xfjc BA eu'deiac xal xfjc T0A Tiepicpepelac aatdarjc yoviac o^etac eu-duypd^tuou ^iei£«v eaxiv, f] Be XoiTif] f] tic- pieXO[Levf] utio xe xfjc T9A Tiepicpepelac xal xfjc AE eu-Mac a7idar]c ywvlac o^eiac eu-duypd^uou eXdxxwv eaxlv. EE yap eaxl xic ywvia euiJuypa^oc [lz'\Z,(J\> [lev xfjc 7iepiexo[ievr)c Otio xe xfjc BA eu'deiac xal xfjc T9A Tiepi- cpepelac, eXdxxtov Be xfjc Tiepiexo[ievr)c utio xe xfjc T0A Tiepicpepelac xal xf)c AE eu'deiac, eic xov ^texacu xotiov xfjc xe rOA Tiepicpepelac xal xfjc AE cu^elac euiJeTa Tiapeu- TieaeTxai, fjxic Tioifjaei jiel^ova [lev xfjc 7iepie)(o^evr)c Otio xe xfjc BA eu'deiac xal xfjc TQA Tiepicpepelac utio eu'deifiv 7iepiexo[ievr)v, eXdxxova Be xfjc Tiepiexo^ievrjc Otco xe xfjc rOA Tiepicpepelac xal xfjc AE eu'deiac. ou Tiape^miTixei Be- oux apa xfjc Tiepiexo^ev7)c ywvlac utio xe xfjc BA eu'deiac xal xfjc rOA Tiepicpepelac eaxai fiel^cov 6$eTa utio eu'deifiv Tiepiexo^evT), ouBe [jurjv eXdxxwv xfjc Tiepiexo^ievr]c Otco xe xfjc T0A Tiepicpepelac xal xfjc AE eu'deiac. to BA, will not fall inside the circle. So, similarly, we can show that neither (will it fall) on the circumference. Thus, (it will fall) outside (the circle) . B Let it fall like AE (in the figure) . So, I say that another straight-line cannot be inserted into the space between the straight-line AE and the circumference CHA. For, if possible, let it be inserted like FA (in the fig- ure), and let DG have been drawn from point D, perpen- dicular to FA [Prop. 1.12]. And since AGD is a right- angle, and DAG (is) less than a right-angle, AD (is) thus greater than DG [Prop. 1.19]. And DA (is) equal to DH. Thus, DH (is) greater than DG, the lesser than the greater. The very thing is impossible. Thus, another straight-line cannot be inserted into the space between the straight-line (AE) and the circumference. And I also say that the semi-circular angle contained by the straight-line BA and the circumference CHA is greater than any acute rectilinear angle whatsoever, and the remaining (angle) contained by the circumference CHA and the straight-line AE is less than any acute rec- tilinear angle whatsoever. For if any rectilinear angle is greater than the (an- gle) contained by the straight-line BA and the circum- ference CHA, or less than the (angle) contained by the circumference CHA and the straight-line AE, then a straight-line can be inserted into the space between the circumference CHA and the straight-line AE — anything which will make (an angle) contained by straight-lines greater than the angle contained by the straight-line BA and the circumference CHA, or less than the (angle) contained by the circumference CHA and the straight- line AE. But (such a straight-line) cannot be inserted. Thus, an acute (angle) contained by straight-lines cannot be greater than the angle contained by the straight-line BA and the circumference CHA, neither (can it be) less than the (angle) contained by the circumference CHA and the straight-line AE. 87 ETOIXEIfiN y'. ELEMENTS BOOK 3 I16pia[Jia. 'Ex 6r] xouxou cpavepov, oxi f] xrj Bia^iexpw toO xuxXou npoq op'&a.z dm' axpa<; &yo(ji£vr] ecpdrcxexai xou xuxXou [xai oxi eO'deTa xuxXou xa$' ev ^xovov scpdrcxexai ar^eiov, £Tiei.8r]Txep xal f] xaxa 5uo auxo au^pdXXouaa svxoi; auxoO TUTtxouaa eSsi/iSr]]- onep eBsi SeT^ai. Arco xou hoftevToc, ay][ieiou xoO Scdsvxoi; xuxXou scpa- uxo^ievTjv euiiteTav ypa^^v dtYaysTv. A "Eaxco xo ^jlsv Bo'dev ar)ueiov xo A, 6 8e Bo'dek; xuxXo<; 6 BrA- 8a Sr) aito xou A ar^dou xoO BrA xuxXou ecpa- 7ixo[ievr)v eu'deTocv Ypa^rjv dyayETv. ElX^cp-dw y^P TO xevxpov xoO xuxXou xo E, xai imZ^X^ ^ AE, xdi xsvxpw [Lev xw E Siaaxrj^taxi 8s tu EA xuxXo<; yeypdcp'dG) 6 AZH, xal duo xou A xfj EA 7tp6<; opMc fj)cdo ^ AZ, xal ensCeux'dwaav ad EZ, AB- Xey«, oxi dmo xou A arjudou xou BrA xuxXou eqxntxouevr] fjxxai f) AB. 'Etcei yap xo E xevxpov eaxl xwv BrA, AZH xuxXwv, Xar] apa eaxlv f] ^isv EA xrj EZ, f] 5e EA xrj EB- Suo 8r) ai AE, EB 8uo xau; ZE, EA taai riaiv xal Y^viav xoivrjv 7i£pi£)(ouai xf|v 7tp6<; xw E- pdaic; apa f\ AZ pdaei xrj AB Xai] saxiv, xal xo AEZ xpiYWvov x£> EBA xpiY^vw laov eaxiv, xal ai Xomal Y^wiai xalc; Xoircau; Y^viaic;- Xor\ apa r\ bub EAZ xrj Otto EBA. opfir} 8e f) Otto EAZ- op-dr) apa xal r) U7i6 EBA. xa[ eaxiv f) EB ex xou xevxpou- f) 8e xrj Bia^texpw xou xuxXou Ttp6<; 6pM<; arc' axpac; ayo[Levr] ecpdnxexai xou xuxXou- f) AB apa ecpaTtxexai xou BrA xuxXou. Atco xou apa So'devxoc; ar)[ieiou xou A xou So-devxoc; xuxXou xou BrA sqjaKxo^svr] eui9sTa Ypa^r] fjxxai f) AB- onep ebei Ttoifjaai. Corollary So, from this, (it is) manifest that a (straight-line) drawn at right-angles to the diameter of a circle, from its extremity, touches the circle [and that the straight-line touches the circle at a single point, inasmuch as it was also shown that a (straight-line) meeting (the circle) at two (points) falls inside it [Prop. 3.2] ]. (Which is) the very thing it was required to show. Proposition 17 To draw a straight-line touching a given circle from a given point. A Let A be the given point, and BCD the given circle. So it is required to draw a straight-line touching circle BCD from point A. For let the center E of the circle have been found [Prop. 3.1], and let AE have been joined. And let (the circle) AFG have been drawn with center E and radius EA. And let DF have been drawn from from (point) D, at right-angles to EA [Prop. 1.11]. And let EF and AB have been joined. I say that the (straight-line) AB has been drawn from point A touching circle BCD. For since E is the center of circles BCD and AFG, EA is thus equal to EF, and ED to EB. So the two (straight-lines) AE, EB are equal to the two (straight- lines) FE, ED (respectively). And they contain a com- mon angle at E. Thus, the base DF is equal to the base AB, and triangle DEF is equal to triangle EBA, and the remaining angles (are equal) to the (corre- sponding) remaining angles [Prop. 1.4]. Thus, (angle) EDF (is) equal to EBA. And EDF (is) a right-angle. Thus, EBA (is) also a right-angle. And EB is a ra- dius. And a (straight-line) drawn at right-angles to the diameter of a circle, from its extremity, touches the circle [Prop. 3.16 corr.]. Thus, AB touches circle BCD. Thus, the straight-line AB has been drawn touching 88 ETOIXEIfiN y'. ELEMENTS BOOK 3 IT]'. 'Edv xuxXou ecpdTCxrjxai tic; eu'deTa, duo 8s xou xevxpou era xr)v 6tcpr]v era£eu)edfi ECcdeia, #] eraCeux'delaa xd-dexoc; eaxai era xrjv ecpa7ixouevr)v. KuxXou ydp xou ABr ecpaTtxecrda) xlc eO^eTa f\ AE xaxd xo T ar)ueTov, xdi eiXr|cp , d« xo xevxpov xou ABr xuxXou xo Z, xal duo xou Z era xo T C7teCeux'0c>) f) ZE Xey«, oxi #] Zr xd^exoc; eaxiv era x/]v AE. El ydp [ir\, r\x&^> duo xou Z era xrjv AE xd-dexoc; f\ ZH. 'Eitel oGv f) Grab ZHr ywvia op'drj eaxiv, o^ela dpa eaxlv f] (mo ZrH- (mo 5e xr|v [ieiCova yaviav f] ^eiX«v uXeupd (moxeivei- ueiCwv dpa f] Zr xfj? ZH- lor\ 5e f] Zr xfj ZB- [ie'i^v dpa xal f] ZB xfjc ZH f] eXdxxwv xrjc; [lei^ovoc;- oracp eaxlv dBuvaxov. oux dpa f] ZH xd-dexoc; eaxiv era xr|v AE. o^ioiwg 5r] SeT^ouev, oxi ou8' aXXr] xi? kX/]v xrjc ZE f] Zr dpa xd-dexoc; eaxiv era x/]v AE. 'Eav dpa xuxXou ecpaTixrjxai tic, euiJeTa, drab Be xou xevxpou era x/]v dcprjv eraCeux'dfj xic; eu-dela, f] eraCeux'deTaa xd-dexoi; eaxai era x/]v ecpaTtxo^ievr]v oraep e5ei BeT^ai. 'Eav xuxXou ecpdn;xr|xai xu; euiSeTa, drab 8e xrjc; dcpfjc; xfj ecpanxoL/evr] 7tp6<; op'ddi; [ycoviac;] eu^eTa yp°W^ &x®% £™- xrjc; dx^eiorjc; eaxai xo xevxpov xou xuxXou. KuxXou yap xou ABr ecpararecrdco xlc; eui&eTa f] AE xaxd xo T arj^ieTov, xal drab xou T xfj AE 7tp6<; op-ddc; f]x'd« r) IA- Xeyco, oxi era xrjc Ar eaxi xo xevxpov xou xuxXou. the given circle BCD from the given point A. (Which is) the very thing it was required to do. Proposition 18 If some straight-line touches a circle, and some (other) straight-line is joined from the center (of the cir- cle) to the point of contact, then the (straight-line) so joined will be perpendicular to the tangent. D For let some straight-line DE touch the circle ABC at point C, and let the center F of circle ABC have been found [Prop. 3.1], and let FC have been joined from F to C. I say that FC is perpendicular to DE. For if not, let FG have been drawn from F, perpen- dicular to DE [Prop. 1.12]. Therefore, since angle FGC is a right-angle, (angle) FCC is thus acute [Prop. 1.17]. And the greater angle is subtended by the greater side [Prop. 1.19]. Thus, FC (is) greater than FG. And FC (is) equal to FB. Thus, FB (is) also greater than FG, the lesser than the greater. The very thing is impossible. Thus, FG is not perpendicular to DE. So, similarly, we can show that neither (is) any other (straight-line) except FC. Thus, FC is perpendicular to DE. Thus, if some straight-line touches a circle, and some (other) straight-line is joined from the center (of the cir- cle) to the point of contact, then the (straight-line) so joined will be perpendicular to the tangent. (Which is) the very thing it was required to show. Proposition 19 If some straight-line touches a circle, and a straight- line is drawn from the point of contact, at right- [angles] to the tangent, then the center (of the circle) will be on the (straight-line) so drawn. For let some straight-line DE touch the circle ABC at point C. And let CA have been drawn from C, at right- 89 ETOIXEIfiN y'. ELEMENTS BOOK 3 angles to DE [Prop, circle is on AC. 1.11]. I say that the center of the Mr] ydp, dXX' et Suvaxov, eaxto to Z, xal sne^sux'dco f\ rz. 'EticI [ouv] xuxXou toO ABr ecpdiixexai tic; euiMa f] AE, duo 6e tou xevxpou era x/]v acp/jv CTiei^euxxai f] Zr, f] Zr dpa xd-dexoc; eaxiv enl xrjv AE- op'&f] dpa eaxlv f) utio ZrE. eaxi 6e xdi #j utio ArE op-dry Tor) dpa eaxlv f) Otto ZrE xfj utio ArE #j eXdxxwv xrj jiei^ovi- oTiep eaxlv dBuvaxov. oux dpa xo Z xevxpov eaxi xoO ABr xuxXou. o^toioc; 8f) 5e(c;o^ev, oxi o08' dXXo xi TtXrjv era xrjc Ar. Edv dpa xuxXou ecpaTixr]xa( tic, eu'deTa, duo 5e xfjc; acpfjc; xfj ecpa7ixo[ievr) Ttpoc; op'ddc; euiDeTa ypa^f) d)fdrj, era xfjc d)(Tf)eiar)c; eaxai xo xevxpov xou xuxXou- ouep e8ei 8eTc;ai. For (if) not, if possible, let F be (the center of the circle), and let CF have been joined. [Therefore], since some straight-line DE touches the circle ABC, and FC has been joined from the center to the point of contact, FC is thus perpendicular to DE [Prop. 3.18]. Thus, FCE is a right-angle. And ACE is also a right-angle. Thus, FCE is equal to ACE, the lesser to the greater. The very thing is impossible. Thus, F is not the center of circle ABC. So, similarly, we can show that neither is any (point) other (than one) on AC. Thus, if some straight-line touches a circle, and a straight- line is drawn from the point of contact, at right-angles to the tangent, then the center (of the circle) will be on the (straight-line) so drawn. (Which is) the very thing it was required to show. x . *Ev xuxXtp f] Tipoc; xfi xevxpw yiovia SiTiXaaitov eaxi xfjc; Ttpoc; xrj Ttepicpepeia, oxav xrjv auxfjv Tiepicpepeiav pdaiv ex«- aiv od ycoviai. 'Eaxco xuxXoc; 6 ABr, xal Tipoc; (iev xw xevxpcp auxou ywvia eaxw f] utio BET, Tipoc; 8e xfj Ttepicpepeia f] utio BAT, exexoaav 8e xf|V auxrjv Tiepicpepeiav pdaiv xrjv Br- Xeyto, oxi SircXacuov eaxlv f) Gtio BET ywvia xfjc; utio BAr. ETiiCeux'delaa yap f] AE Sifix'O" era to Z. 'EticI ouv iar] eaxlv f) EA xfj EB, lay) xal ywvia f) utio EAB xfj utio EBA- ai dpa utio EAB, EBA ycoviai xfjc; utio EAB BiTiXaaiouc; eiaiv. Xor\ 8e f] utio BEZ xau; utio EAB, EBA- xal f] utio BEZ dpa xfjc; utio EAB eaxi 8iTiXfj. Sid xd auxd 8r] xal f] utio ZEE xfjc utio EAT eaxi SiTiXfj. 6Xrj dpa f] utio BEr bXf]z xfjc; utio BAT eaxi SiTiXfj. Proposition 20 In a circle, the angle at the center is double that at the circumference, when the angles have the same circumfer- ence base. Let ABC be a circle, and let BEC be an angle at its center, and BAG (one) at (its) circumference. And let them have the same circumference base BC. I say that angle BEC is double (angle) BAG. For being joined, let AE have been drawn through to F. Therefore, since EA is equal to EB, angle EAB (is) also equal to EBA [Prop. 1.5]. Thus, angle EAB and EBA is double (angle) EAB. And BEF (is) equal to EAB and EBA [Prop. 1.32]. Thus, BEF is also double EAB. So, for the same (reasons), FEC is also double EAC. Thus, the whole (angle) BEC is double the whole (angle) BAG. 90 ETOIXEIfiN y'. ELEMENTS BOOK 3 r B KexXdcrdo) 5r] TidXiv, xai eaxco exepa ywvia f] Otto BAr, xai SKi^eux'Se'iCTa #] AE expepXf]a , dco era to H. 6[ioi«<; 8f) 8ei^o[i£v, oxi BiTiXfj eaxiv f) utto HEr ycovia xfjc; utto EAr, fiv f) Otto HEB SiTcXfj eaxi xfjc; utto EAB- Xomf) apa f) Otto BEr SiTtXfj saxi xfjc; utto BAr. 'Ev xuxXw apa f\ upoc; xw xevxpa) ycovia SmXaaiwv saxi xfjc; Tipoc; xfj Tcepicpepda, oxav xfjv auxf]v Ttepicpepeiav pdaiv S)(coaiv [ai ywviai]- oicsp eSei 8eTc;ai. xa'. 'Ev xuxXip ai ev xO auxcp x[if]|jiaxi ywviai iaai dXXfjXaic; eiaiv. Tiaxw xuxXoc; 6 ABrA, xal ev xw auxfi x^uaxi iu BAEA yioviai eaxwaav ai Otto BAA, BEA- Xeyw, oxi ai Otto BAA, BEA ytoviai loan dXXfjXaic; eiaiv. EiXr]cp'd« yap xou ABrA xuxXou xo xevxpov, xal eaxco xo Z, xai eTteCeuyjdwaav ai BZ, ZA. Kai eicsi f) [lev Otto BZA yiovia Ttpoc; xai xevxpa> eaxiv, f) 8e Otto BAA Ttpoc; xfj Ttepicpepeia, xai exouai T1 ^ v aUT ^ v Tie- picpepeiav pdaiv xfjv BrA, f] apa utto BZA ya>via SiTtXaaiwv eaxi xfjc; Otto BAA. 8ia xa auxa 8f) f) utto BZA xai xfjc; utto c B So let another (straight-line) have been inflected, and let there be another angle, BDC. And DE being joined, let it have been produced to G. So, similarly, we can show that angle GEC is double EDC, of which GEB is double EDB. Thus, the remaining (angle) BEC is double the (remaining angle) BDC. Thus, in a circle, the angle at the center is double that at the circumference, when [the angles] have the same circumference base. (Which is) the very thing it was re- quired to show. Proposition 21 In a circle, angles in the same segment are equal to one another. Let ABCD be a circle, and let BAD and BED be angles in the same segment BAED. I say that angles BAD and BED are equal to one another. For let the center of circle ABCD have been found [Prop. 3.1], and let it be (at point) F . And let BF and FD have been joined. And since angle BFD is at the center, and BAD at the circumference, and they have the same circumference base BCD, angle BFD is thus double BAD [Prop. 3.20]. 91 ETOIXEIfiN y'. ELEMENTS BOOK 3 BE A eaxi 8mXot<ov Xar\ dpa f) utio BAA xfj utio BE A. 'Ev xuxXtp apa al ev xfi auxw x^yjuaxi ytovtcxi i'aai dXXr]Xaic eta£v OTiep eSei 8el^ai. 'Eaxco xuxXo<; 6 ABrA, xal ev auxw xexpdiiXeupov eaxo xo ABrA- Xeyw, oxi at diievavxiov yoviai 8uaiv op'dau; i'aai eiaiv. TkeCeux'dwaav ai Ar, BA. 'Etiei ouv navxoc; xpiycovou ai xpeu; ycoviai 8uaiv opiDaTt; laai eiaiv, xoO ABr apa xpiywvou ai xpeu; ycoviai ai utio TAB, ABr, BrA 8uaiv opiJau; i'aai eiaiv. Xor\ 8s t\ fiev utio TAB xfj uko BAr- ev yap ic5 auxw xjir^axi eiai xo BAAT- f] 8s utio ATB xfj utio AAB- ev yap xc5 auxo x^xr^axi eiai xo AArB- 6Xr) apa f] utio AAr xau; Otio BAr, ArB 'ior\ eaxiv. xoivf) Tipoaxeio^w f) utio ABr- ai apa utio ABr, BAr, ArB xau; utio ABr, AAr i'aai eiaiv. dXX'' ai utio ABr, BAr, ArB Suaiv bpftdXc, i'aai eiaiv. xai ai utio ABr, AAr dpa 8uaiv op'daic; laai eiaiv. b\ioicdz 8r) Sei^o^tev, oxi xai ai utio BAA, ArB ywviai Suaiv opiJau; i'aai eiaiv. Ttov apa ev xou; xuxXou; xexpaTiXeupwv ai diievavxiov ycoviai 8uaiv op'ddu; laai eiaiv oTiep e8ei Sel^ai. xy'. 'Era xfjc auxfjc eMeiac 8uo x|ir|jiaxa xuxXcov o^ioia xai dviaa ou auaxaiDrpexai era xd auxa {iepr\. Ei yap 8uvax6v, era xrj<; auxrj<; euTDeia<; xfj<; AB Suo x^ir][iaxa xuxXwv ojioia xai dviaa auveaxdxw era xd auxa [iepr] xd ArB, AAB, xai 8ir])fdo f] ATA, xai eTre^eu)cd«aav So, for the same (reasons), BFD is also double BED. Thus, BAD (is) equal to BED. Thus, in a circle, angles in the same segment are equal to one another. (Which is) the very thing it was required to show. Proposition 22 For quadrilaterals within circles, the (sum of the) op- posite angles is equal to two right-angles. Let ABCD be a circle, and let ABCD be a quadrilat- eral within it. I say that the (sum of the) opposite angles is equal to two right-angles. Let AC and BD have been joined. Therefore, since the three angles of any triangle are equal to two right-angles [Prop. 1.32], the three angles CAB, ABC, and BCA of triangle ABC are thus equal to two right-angles. And CAB (is) equal to BDC. For they are in the same segment BADC [Prop. 3.21]. And ACB (is equal) to ADB. For they are in the same seg- ment ADCB [Prop. 3.21]. Thus, the whole of ADC is equal to BAC and ACB. Let ABC have been added to both. Thus, ABC, BAC, and ACB are equal to ABC and ADC. But, ABC, BAC, and ACB are equal to two right-angles. Thus, ABC and ADC are also equal to two right-angles. Similarly, we can show that angles BAD and DCB are also equal to two right-angles. Thus, for quadrilaterals within circles, the (sum of the) opposite angles is equal to two right-angles. (Which is) the very thing it was required to show. Proposition 23 Two similar and unequal segments of circles cannot be constructed on the same side of the same straight-line. For, if possible, let the two similar and unequal seg- ments of circles, ACB and ADB, have been constructed on the same side of the same straight-line AB. And let 92 ETOIXEIfiN y'. ELEMENTS BOOK 3 ai TB, AB. 'Ercei ouv 6\loi6v eaxi to ArB xji/jjia iu AAB x^^axi, ojjioia 8e xjir^iaxa xuxAiov eaxl xd 8e)(6[ieva y^viac; laac, Tar) apa eaxlv f) uno ArB ycovia xfj Otio AAB f) exxoc; xrj evxoc;- onep eaxlv d8uvaxov. Oux apa em xrjc; auxrjc; euiSeiac; 860 x|ir][iaxa xuxXwv o|ioia xal aviaa auaxai3r]0£xai era xa auxd y.epi]- onep eSei 8eTc;ai. ^4CZ? have been drawn through (the segments), and let CB and DB have been joined. A B Therefore, since segment ACB is similar to segment ADB, and similar segments of circles are those accept- ing equal angles [Def. 3.11], angle ACB is thus equal to ADB, the external to the internal. The very thing is impossible [Prop. 1.16]. Thus, two similar and unequal segments of circles cannot be constructed on the same side of the same straight-line. x5'. Proposition 24 Td em '(awv euiJeiMv o|ioia x^uaxa xuXgjv I'oa aXki]ko\.c, eaxiv. Similar segments of circles on equal straight-lines are equal to one another. E r A "Eaxwaav yap era lawv eu-deiwv x£Sv AB, TA o^ioia x^tr]uaxa xuxXwv xa AEB, TZA- Xeyw, oxi I'aov eaxl xo AEB x^ifjjia t& TZA x[if]^axi. 'Ecpap|jio^o|jievou yap xou AEB x^tr]uaxo<; era xo TZA xal xidejievou xou [ie\ A o/jjieiou era xo T xrjc; 8e AB euiSetai; era xr]v TA, ecpapjioaei xal xo B o/jjielov era xo A arj^ielov Sid xo Xat]\i elvai xrjv AB xfj TA- xfjc 8e AB era xrjv TA ecpap- y.oc6>ar\c, ecpap^iooei xal xo AEB xjifj^ia era xo TZA. ei yap f) AB eu'dela era xrjv TA ecpap^xoaei, xo 8e AEB xjrrj^a era xo TZA jit) ecpapjiooei, fjxoi evxoc; auxou neaelxai f\ exxoc; f] n:apaXXdc;ei, d>c; xo THA, xal xuxXoc; xuxXov xejivei xaxd TiXeiova cnqjiela fj 8uo' onep eaxiv dBuvaxov. oux apa ecpap- [ioZ,oy.evf]z xrjc; AB eO'deiac; era xrjv TA oux ecpapjiooei xal C D For let AEB and CFD be similar segments of circles on the equal straight-lines AB and CD (respectively) . I say that segment AEB is equal to segment CFD. For if the segment AEB is applied to the segment CFD, and point A is placed on (point) C, and the straight-line AB on CD, then point B will also coincide with point D, on account of AB being equal to CD. And if AB coincides with CD then the segment AEB will also coincide with CFD. For if the straight-line AB coincides with CD, and the segment AEB does not coincide with CFD, then it will surely either fall inside it, outside (it),^ or it will miss like CCD (in the figure), and a circle (will) cut (another) circle at more than two points. The very 93 ETOIXEIfiN y'. ELEMENTS BOOK 3 to AEB Turj^ia era to TZA- ecpapjioaei apa, xal taov ai)T« eaTai. Td apa era lacov eO'deioov o^ioia T^f|^aTa xuxXwv laa dXXyjXoic; ecrav oTcep eBei 5eT?ai. t Both this possibility, and the previous one, are precluded by Prop. 3.23. xs'. KuxXou x^jiaxoc; So-devxoc; Tcpoaavaypdcj;ai tov xuxXov, ounep eaxi Tjxfj^a. r r r "Eotw to 5oi9ev xurjpc xuxXou to ABE 8el Bf] toO ABr xp^axoc; Ttpoaavaypdcjjai tov xuxXov, ouTcep eaxi xjirjjia. Tex[i rjcrdco yap f] Ar Stya xaxd to A, xal rjx'dco duo xou A arjueiou xfj Ar Tcpoc; opMc; f] AB, xal erav^eux'&co i) AB- f] Otco ABA yovla apa xrjc; Otto BAA f]xoi ^elc^cov eaxlv rj tar] fj eXdxxwv. "Eaxco TcpoTepov iiei^cov, xal auveoxdxo repot; xfj BA eO'dela xal x£> Tcpoc; aOxfj ar^eio tw A Tfj Otco ABA ywvia Xor\ f) Otco BAE, xal 8ir])fdw f] AB era to E, xal erceCeux'dco f] Er. excel ouv lay] eaxlv rj Otco ABE ycovia xfj Otco BAE, lay] apa eaxi xal y) EB eu-dela Tfj EA. xal ercel iar] eaxlv f) AA Tfj Ar, xoivf] Se r) AE, Suo Bf] al AA, AE Buo Talc; TA, AE laai eialv exaxepa exaxepa- xal ycovia y) Otco AAE ycovia xfj Otco TAE eaxiv car)" dp'df) yap exaTepa' pdaic; apa f] AE pdoei Tfj TE eaxiv car). dXXd y) AE Tfj BE eSd^r] lay]- xal f) BE apa Tfj TE eaxiv Iar)- ai xpelc; apa ai AE, EB, Er I'aai dXXf|Xaic; eia(v 6 apa xevxpaS x£> E 8iaaxf]^axi Be evl twv AE, EB, Er xuxXoc; ypacpouevoc; f^ei xal Sid xwv Xoitcwv ar)[ieiwv xal eaxai Tcpoaavayeypa^evoc;. xuxXou apa x^if^axoc; So'devxoc; TtpoaavayeypaTcxai 6 xuxXog. xal 8fjXov, cbc; to ABr T^ifjiia eXaxxov eaxiv f]^ixuxX(ou Sid to to E xevxpov 6xt6c; auxou Tuy)(dveiv. 'Oiioiwc; [8e] xav f) f) Otco ABA ycovia lay] Tfj Otco BAA, xfjc; AA larf, yevo|ievr)<; exaTepa xoov BA, Ar ai Tpelc; al AA, AB, Ar Taai dXXfjXaic; eaovxai, xal eaxai to A xevxpov toO TcpoaavaTceTcXr)pw^ievou xuxXou, xal SrjXaSf) eaxai to ABr fjuixuxXiov. 'Edv Be f) Otco ABA eXdxxwv fj xfjc; Otco BAA, xal au- axrjao^cda Tcpoc; Tfj BA euiJeia xal iS Tcpoc; auxfj ar^elw thing is impossible [Prop. 3.10]. Thus, if the straight-line AB is applied to CD, the segment AEB cannot not also coincide with CFD. Thus, it will coincide, and will be equal to it [C.N. 4]. Thus, similar segments of circles on equal straight- lines are equal to one another. (Which is) the very thing it was required to show. Proposition 25 For a given segment of a circle, to complete the circle, the very one of which it is a segment. A A . A Let ABC be the given segment of a circle. So it is re- quired to complete the circle for segment ABC, the very one of which it is a segment. For let AC have been cut in half at (point) D [Prop. 1.10], and let DB have been drawn from point D, at right-angles to AC [Prop. 1.11]. And let AB have been joined. Thus, angle ABD is surely either greater than, equal to, or less than (angle) BAD. First of all, let it be greater. And let (angle) BAE, equal to angle ABD, have been constructed on the straight-line BA, at the point A on it [Prop. 1.23]. And let DB have been drawn through to E, and let EC have been joined. Therefore, since angle ABE is equal to BAE, the straight-line EB is thus also equal to EA [Prop. 1.6]. And since AD is equal to DC, and DE (is) common, the two (straight-lines) AD, DE are equal to the two (straight-lines) CD, DE, respectively. And angle ADE is equal to angle CDE. For each (is) a right-angle. Thus, the base AE is equal to the base CE [Prop. 1.4]. But, AE was shown (to be) equal to BE. Thus, BE is also equal to CE. Thus, the three (straight-lines) AE, EB, and EC are equal to one another. Thus, if a cir- cle is drawn with center E, and radius one of AE, EB, or EC, it will also go through the remaining points (of the segment), and the (associated circle) will have been completed [Prop. 3.9]. Thus, a circle has been completed from the given segment of a circle. And (it is) clear that the segment ABC is less than a semi-circle, because the center E happens to lie outside it. 94 ETOIXEIfiN y'. ELEMENTS BOOK 3 tu A xfj Grab ABA yovia larjv, evxo<; xou ABr x^rjpcxoc; TteaeTxai xo xevxpov era xrjg AB, xal eaxai SrjXaBr] xo ABr xjji/j^ia ^iel£ov fjjiixuxXiou. KuxXou apa x^fj^axoc; 8oiL)evxo<; TtpoaavayeypaTixai ° xuxXog- onep eSei Hoifjaai. [And] , similarly, even if angle ABD is equal to BAD, (since) AD becomes equal to each of BD [Prop. 1.6] and DC, the three (straight-lines) DA, DB, and DC will be equal to one another. And point D will be the center of the completed circle. And ABC will manifestly be a semi-circle. And if ABD is less than BAD, and we construct (an- gle BAE), equal to angle ABD, on the straight-line BA, at the point A on it [Prop. 1.23], then the center will fall on DB, inside the segment ABC. And segment ABC will manifestly be greater than a semi-circle. Thus, a circle has been completed from the given seg- ment of a circle. (Which is) the very thing it was required to do. X9 Proposition 26 'Ev xolc; iaoiz xuxXou; ai Xaai ycoviai era i'awv Ttepicpe- peiwv pcprjxaaiv, eav xe 7ip6<; xoTc, xevxpou; eav xe npbq xau; Ttspicpepdocu; Sai pepr]X\iiai. A. K A "Eaxcoaav Taoi xuxXoi ol ABr, AEZ xal ev auxou; Taai ycoviai eoxwaav upoc; \xk\ xou; xevxpou; ai bub BHr, E9Z, Ttpoc; Se xau; Ttepicpepeiau; ai Gtio BAr, EAZ- Xeyco, oxi Tar) eaxlv f| BKr rcepicpepeia xfj EAZ nepicpepeia. 'Erac^e6)cd«oav yap ai Br, EZ. Kai CTiel Taoi eialv oi ABr, AEZ xuxXoi, Taai eialv ai ex xfiv xevxptov 8uo 8r] ai BH, Hr 60o xau; E9, 9Z Taai- xal yiovia f\ npbc, xai H ywvia xfj 7ip6c; xai O Tar)- pdau; dpa f] Br pdoei xfj EZ eaxiv i'ar). xal inel Tar] eaxlv rj Ttpoc; iw A ywvia xfj npbz x« A, ojioiov dpa eaxl xo BAr x^rj^ia xG EAZ xjir^iaxi- xai eioiv era Tatov euiSeiSv [xfiv Br, EZ]' xa 8e era Tacov eMeifiv ojioia xjir^axa xuxXtov Taa dXXrjXou; eaxiv Taov apa xo BAr xjrrjua xG EAZ. eaxi 8e xal oXoc; 6 ABr xuxXoc; oXw ifi AEZ xuxXtp Taoc;- Xoinr) apa f) BKr Ttepicpepeia xfj EAZ Ttepicpepeia eaxlv i'ar). 'Ev apa xou; Taou; xuxXoic; ai i'aai yioviai era Taa>v Ttepi- cpepeifiv peprjxaaiv, edv xe icpoc; xou; xevxpou; eav xe icpoc; xau; uepicpepeiac; fiai peprjxuTai- onep e8ei SeT^ai. In equal circles, equal angles stand upon equal cir- cumferences whether they are standing at the center or at the circumference. K L Let ABC and DEF be equal circles, and within them let BGC and EHF be equal angles at the center, and BAG and EDF (equal angles) at the circumference. I say that circumference BKC is equal to circumference ELF. For let BC and EF have been joined. And since circles ABC and DEF are equal, their radii are equal. So the two (straight-lines) BG, GC (are) equal to the two (straight-lines) EH, HF (respectively). And the angle at G (is) equal to the angle at H. Thus, the base BC is equal to the base EF [Prop. 1.4]. And since the angle at A is equal to the (angle) at D, the segment BAC is thus similar to the segment EDF [Dei. 3.11]. And they are on equal straight-lines [BC and EF] . And simi- lar segments of circles on equal straight-lines are equal to one another [Prop. 3.24]. Thus, segment BAC is equal to (segment) EDF. And the whole circle ABC is also equal to the whole circle DEF. Thus, the remaining circum- ference BKC is equal to the (remaining) circumference ELF. Thus, in equal circles, equal angles stand upon equal circumferences, whether they are standing at the center 95 ETOIXEIfiN y'. ELEMENTS BOOK 3 'Ev xou; laoic; xuxXoig al era lacov Tiepicpepeuov pepiqxuTai ycoviai i'aai dXXf|Xai<; eiaiv, edv xe Tipoc; xou; xevxpou; edv xe Tipoc; xau; Tiepicpepeiau; Sai pe(3r)xuTai. ^ A or at the circumference. (Which is) the very thing which it was required to show. Proposition 27 In equal circles, angles standing upon equal circum- ferences are equal to one another, whether they are standing at the center or at the circumference. A_ _ n 'Ev yap looic, xuxXoig xou; ABr, AEZ Ira Tacov Tiepi- cpepeifiv iwv Br, EZ Tipoc; y.sv xolc; H, xevxpou; ycoviai PePrjxexcoaav ai utio BBT, E9Z, Tipoc; 8e xau; Tiepicpepeiau; ai utio BAr, EAZ- Xeyw, oxi f) |iev utio BBT ywvia xfj utio E9Z eaxiv larj, f) Se utio BAr xfj utio EAZ eaxiv for). Et yap dviaoc; eaxiv f] utio BHr xfj 6ti6 E9Z, [iia auxGv ^iei£«v eaxiv. eaxw ^eic^cov f] utio BHr, xod auveaxdxco Tipoc; xfj BH eu'deia xal iu Tipoc; auxfj ar^eicp x6> H xfj Otio E9Z ywvia for) f) Otio BHK- ai Se foai yoviai era lowv TiepicpepeiGv pepf|xaaiv, oxav Tipoc; xou; xevxpou; Saiv Tar] dpa f] BK TxepLcpepeia xfj EZ Tiepicpepeia. dXXd f] EZ xfj Br eaxiv for)- xal f) BK dpa xfj Br eaxiv for] f] eXdxxtov xfj \ieiZovv oTiep eaxiv dSuvaxov. oux dpa dviaoc; eaxiv f) utio BHr ytovia xfj utio E9Z- for] dpa. xai eaxi xfjc; [lev utio BHr fpiaeia f] Tipoc; xw A, xrjc Se utio E9Z fjuiaeia f) Tipoc; to A- for] dpa xal f] Tipoc; xw A yovia xfj Tipoc xw A. 'Ev dpa xou; foou; xuxXou; ai era i'awv Tiepicpepeiov pe- Pr]XuTai ywviai i'aai dXXf|Xai<; eiaiv, edv xe Tipoc; xou; xevxpou; edv xe Tipoc talc Tiepicpepeiau; CSai pepr]xuiai- oTiep e8ei SeT^ai. XT] . 'Ev xou; foou; xuxXou; ai i'aai eui5elai i'aac; Tiepicpepeiac; dcpaipoOai xf]v (iev (iei^ova xfj jiei^ovi xfjv Se eXdxxova xfj eXdxxovi. 'Eaxwaav laoi xuxXoi oi ABr, AEZ, xai ev xou; xuxXou; foai eu^elai eaxcoaav ai AB, AE xdc; [ie\ ATB, AZE Tiepi- cpepeiac; jiei^ovac; dcpaipouaai xdc; Se AHB, A9E eXdxxovag- Xeyoj, oxi f] [iev ArB jiei^wv Tiepicpepeia for] eaxi xfj AZE ^ei^ovi Tiepicpepeia f) 8e AHB eXdxxtov Tiepicpepeia xfj A9E. For let the angles BGC and EHF at the centers G and H, and the (angles) BAG and EDF at the circum- ferences, stand upon the equal circumferences BC and EF, in the equal circles ABC and DEF (respectively) . I say that angle BGC is equal to (angle) EHF, and BAC is equal to EDF. For if BGC is unequal to EHF, one of them is greater. Let BGC be greater, and let the (angle) BGK, equal to angle EHF, have been constructed on the straight-line BG, at the point G on it [Prop. 1.23]. But equal angles (in equal circles) stand upon equal circumferences, when they are at the centers [Prop. 3.26]. Thus, circumference BK (is) equal to circumference EF. But, EF is equal to BC. Thus, BK is also equal to BC, the lesser to the greater. The very thing is impossible. Thus, angle BGC is not unequal to EHF. Thus, (it is) equal. And the (angle) at A is half BGC, and the (angle) at D half EHF [Prop. 3.20]. Thus, the angle at A (is) also equal to the (angle) at D. Thus, in equal circles, angles standing upon equal cir- cumferences are equal to one another, whether they are standing at the center or at the circumference. (Which is) the very thing it was required to show. Proposition 28 In equal circles, equal straight-lines cut off equal cir- cumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser. Let ABC and DEF be equal circles, and let AB and DE be equal straight-lines in these circles, cutting off the greater circumferences ACB and DFE, and the lesser (circumferences) AGB and DHE (respectively). I say that the greater circumference ACB is equal to the greater circumference DFE, and the lesser circumfer- 96 ETOIXEIfiN y'. ELEMENTS BOOK 3 r Z H © ElXricp-dw ydp T °< xfvxpa iSv xuxXwv xa K, A, xal sriie^Eux'dwaav ad AK, KB, AA, AE. Kcd inei Taoi xuxXoi eiaiv, Taai rial xal ai sx xfiv xsvxpcov 860 8r) ai AK, KB Sua! xdi<; AA, AE Taai eiaiv xal |3daic; f) AB pdoei xfj AE Tory ywvia apa f) Otto AKB ycovia xfj Otto AAE Tar] eaxiv. ai 8s Taai ycoviai km. Tacov Tispicpspeiwv pepf]xaaiv, oxav npoc, xoT? xevxpoi? Saiv Tar) apa f] AHB nepicpepsia xfj AGE. saxl 6e xai okoc, 6 ABr xuxXog 6Xcp xcp AEZ xuxXcp Taog- xal Xomf] apa f) ArB nepicpepsia Xomfj xfj AZE rcspicpspeia i'or) eaxiv. 'Ev apa xolc; Taoic; xuxXoic; ai Taai euiSriai Taac; m- picpspeiac; dcpaipouai xf]v (lev jiei^ova xrj jirii^ovi xf]v 8e sXdxxova xrj eXdxxovi- onep e8si SeT^ai. ence AGB to (the lesser) DHE. C F G H For let the centers of the circles, K and L, have been found [Prop. 3.1], and let AK, KB, DL, and LE have been joined. And since {ABC and DEF) are equal circles, their radii are also equal [Def. 3.1]. So the two (straight- lines) AK, KB are equal to the two (straight-lines) DL, LE (respectively). And the base AB (is) equal to the base DE. Thus, angle AKB is equal to angle DLE [Prop. 1.8]. And equal angles stand upon equal circum- ferences, when they are at the centers [Prop. 3.26] . Thus, circumference AGB (is) equal to DHE. And the whole circle ABC is also equal to the whole circle DEF. Thus, the remaining circumference ACB is also equal to the remaining circumference DFE. Thus, in equal circles, equal straight-lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser. (Which is) the very thing it was required to show 'Ev xoi<; Xaoic, xuxXou; xdc; Taac; Ttepicpepriac; Taai cu-deiai uitoxrivouaiv. A A H "Eaxcoaav i'aoi xuxXoi oi ABr, AEZ, xai ev auxdic; Taai Ttepicpepsiai a7t£iXf]cpiL>Gjaav ai BHr, E0Z, xal ErcriCeuyjiEkoaav ai Br, EZ su'deTai- Xeyio, oxi Tar] eaxiv f] Br xfj EZ. EiXr'icp'dw yap xa xevxpa xGv xuxXwv, xai eaxw xd K, A, xai STieCeux'dwaav ai BK, KT, EA, AZ. Kai ETisiTaT) eaxiv f] BHr Ttepicpepeia xfj E9Z Ttepicpepeia, Proposition 29 In equal circles, equal straight-lines subtend equal cir- cumferences. A D G H Let ABC and DEF be equal circles, and within them let the equal circumferences BGC and EHF have been cut off. And let the straight-lines BC and EF have been joined. I say that BC is equal to EF. For let the centers of the circles have been found [Prop. 3.1], and let them be (at) K and L. And let BK, 97 ETOIXEIfiN y'. ELEMENTS BOOK 3 Tor) eoxl xal yoivia f\ uno BKr xrj Otto EAZ. xocl tizzi 1001 eiolv oi ABr, AEZ xuxXoi, Toai rial xal al ex xc5v xevxptov 860 8r) ai BK, Kr Bual xalc; EA, AZ I'oai eiolv xal ytovtag Ioa<; Tiepiexouaiv pdoi<; apa f) Br pdoei xfj EZ I'or) eoxiv 'Ev apa xolc Took; xuxXok; xac laac, Ttepicpepeiac; foai eu'deTai utcoxcivouoiv ousp e8ei Ssl^ai. KC, EL, and LF have been joined. And since the circumference BGC is equal to the cir- cumference EHF, the angle BKC is also equal to (an- gle) ELF [Prop. 3.27]. And since the circles ABC and DEF are equal, their radii are also equal [Def. 3.1]. So the two (straight-lines) BK, KC are equal to the two (straight-lines) EL, LF (respectively) . And they contain equal angles. Thus, the base BC is equal to the base EF [Prop. 1.4]. Thus, in equal circles, equal straight-lines subtend equal circumferences. (Which is) the very thing it was required to show. X'. Trjv BodeToav rcepicpepeiav Blxa xejielv. A a r b TCaxco f\ Sodriaa rcepicpepeia f] AAB- 8eT 8rj xrjv AAB Ttepicpepeiav 8()(a xefielv. 'Etie^eux^co f] AB, xal xsxjjirjcrdo hiyv. xaxa xo V, xal dno xou T a/jjisbu xrj AB eui&eia rcp6<; op-ddi; f])fdw r) TA, xal £7TE^£U)cdtoaav al AA, AB. Kal en el lay) eoxlv f] AT xfj TB, xoivr] 8e f] TA, 860 8/) ai AT, TA 8uai xalg Br, TA I'oai eloiv xal ytovia f] utt:6 ArA ywvla xrj (mo BrA far)- op-Qr] yap exaxepa- pdau; apa f] AA pdoei xrj AB Xar\ eoxiv. al 8s: I'oai eO'delai laac; Ttepicpepeiag dcpaipouai xr]v [ie\ jieii^ova xfj ^el^ovi xrjv 8e eXdxxova xrj eXdxxovi- xdi eoxiv exaxepa x«v AA, AB ize- pi9£pei£3v eXdxxcov r^ixuxXlou- I'or) apa f] AA Ttepicpepeia xfj AB nepicpepela. ; H apa SoOelaa nepicpepeia 8i/a xex^rjxai xaxa xo A ar][ieiov ojiep eBei Ttoirjoai. Proposition 30 To cut a given circumference in half. D A C B Let ADB be the given circumference. So it is required to cut circumference ADB in half. Let AB have been joined, and let it have been cut in half at (point) C [Prop. 1.10]. And let CD have been drawn from point C, at right-angles to AB [Prop. 1.11]. And let AD, and DB have been joined. And since AC is equal to CB, and CD (is) com- mon, the two (straight-lines) AC, CD are equal to the two (straight-lines) BC, CD (respectively). And angle ACD (is) equal to angle BCD. For (they are) each right- angles. Thus, the base AD is equal to the base DB [Prop. 1.4]. And equal straight-lines cut off equal circum- ferences, the greater (circumference being equal) to the greater, and the lesser to the lesser [Prop. 1.28]. And the circumferences AD and DB are each less than a semi- circle. Thus, circumference AD (is) equal to circumfer- ence DB. Thus, the given circumference has been cut in half at point D. (Which is) the very thing it was required to do. Xa'. 'Ev xuxXcp f] ^iev ev iu f)puxuxX[tp yavia op-Qr] eoxiv, f] 8e ev xfi ^.eii^ovi x[if]^.axi eXdxxwv op'Sr^, f\ 8e ev xai eXdxxovi x^ir^axi jie^wv op'drjc xal era f] ^iev xoO ]j.ei(l,o\)oc, xur]pcxo<; Y«v(a (j.elC"v eoxlv op'drjc;, f) 8e xou eXdxxovo<; xur]pcxo<; ycovia eXdxxwv op'drjc;. Proposition 31 In a circle, the angle in a semi-circle is a right-angle, and that in a greater segment (is) less than a right-angle, and that in a lesser segment (is) greater than a right- angle. And, further, the angle of a segment greater (than a semi-circle) is greater than a right-angle, and the an- 98 ETOIXEIfiN y'. ELEMENTS BOOK 3 Tiaxw xuxXo<; 6 ABrA, 8id|iexpoc; 8e auxoO eaxto f] BT, xevxpov 8e to E, xal CTce£eu)edk>aav ai BA, Ar, AA, AE Xeyw, oxi f] [iev ev xfi BAr r^ixuxXup yovia f] utco BAr opflVj eaxiv, f] 8e ev xw ABr jieiCovi xoO rpixuxXiou x^r^axi y«v(a f) utco ABr eXdxxcov eaxiv opiDrjc;, f] 8e ev xw AAr eXdxxovi xou f^ixuxXiou x^ir|uaxi ywvia rj Otxo AAr jiei^wv eaxiv op-df^. 'ETieCeu)fdw f) AE, xal Sir^o rj BA era xo Z. Kal CTcei iar] eaxiv f) BE xfj EA, Xar\ eaxl xal ywvia f) utco ABE xfj utco BAE. TcdXiv, enel lor\ eaxiv f] TE xfj EA, iar] eaxl xal f] utco ArE xfj utco TAE- 6Xr] dpa f) Otco BAr Suai xdic utio ABr, ArB Xor\ eaxiv. eaxl 8e xal f] Otco ZAr exxoc xou ABr xpiyovou 8ual xdig Otco ABr, ArB ycoviau; iar)- Iar] dpa xal f] utco BAr y«v[a xfj utco ZAr- op-df] dpa exaxepa- f] dpa ev xw BAr f^ixuxXitp ywvia f] utio BAr opi^f] eaxiv. Kal CTcei xou ABr xpiywvou 8uo ywviai ai utio ABr, BAr 8uo op-dSv eXdxxove<; eiaiv, opftr} 8e f] utco BAr, eXdxxov dpa op-drjc eaxiv f) utco ABr ywvia- xai eaxiv ev x<3 ABr [xeiCovi xou fpixuxXiou xuf^axi. Kal CTiel ev xuxXw xexpaTcXeupov eaxi xo ABrA, xCSv Se ev xou; xuxXok; xexpaTcXeupwv ai dTcevavxiov ycoviai Suaiv opiJau; iaai eiaiv [ai dpa utco ABr, AAr ytoviai 8uaiv opiJau; Taa<; eiaiv], xai eaxiv f] utio ABr eXdxxwv opiDrjc' Xoitct) dpa f] utco AAr ywvia ^tei^Mv op'drjc; eaxiv xai eaxiv ev iu AAr eXdxxovi xou f)(iixuxX[ou x(if|iJiaxi. Aeyw, oxi xai f] (jiev xou x^if|^axo<; ywvia f] Tte- piexo^evr] utco [xe] xrjc; ABr Ttepicpepeiac xai xfj? Ar eu-deiac; [leiCwv eaxiv op'drjc, f] Se xou eXdxxovoc; x^ir](iaxoc; ywvia f\ Tcepiexofievr) utco [xe] xrjc; AA[r] Tcepicpepeiac; xal xfj? Ar eu-deiac; eXdxxwv eaxiv op'drjc;. xa( eaxiv auxo'dev cpavepov. CTcel yap f] utio twv BA, Ar eu-deiwv op'df] eaxiv, f] dpa utco xrjc; ABr Tcepicpepeiac; xal xrjc; Ar eui9eia<; Ttepiexo^tevr] ^ei£tov eaxiv op'drjc;. TidXiv, CTcel f] utco xwv Ar, AZ euTJeiSv op'dr] eaxiv, f] dpa utco xrjc; IA eui!)eia<; xai xrjc; AA[r] rcepi- gle of a segment less (than a semi-circle) is less than a right-angle. Let ABCD be a circle, and let BC be its diameter, and E its center. And let BA, AC, AD, and DC have been joined. I say that the angle BAG in the semi-circle BAC is a right-angle, and the angle ABC in the segment ABC, (which is) greater than a semi-circle, is less than a right- angle, and the angle ADC in the segment ADC, (which is) less than a semi-circle, is greater than a right-angle. Let AE have been joined, and let BA have been drawn through to F. And since BE is equal to EA, angle ABE is also equal to BAE [Prop. 1.5]. Again, since CE is equal to EA, ACE is also equal to CAE [Prop. 1.5]. Thus, the whole (angle) BAC is equal to the two (angles) ABC and ACB. And FAC, (which is) external to triangle ABC, is also equal to the two angles ABC and ACB [Prop. 1.32]. Thus, angle BAC (is) also equal to FAC. Thus, (they are) each right-angles. [Def. 1.10]. Thus, the angle BAC in the semi-circle BAC is a right-angle. And since the two angles ABC and BAC of trian- gle ABC are less than two right-angles [Prop. 1.17], and BAC is a right-angle, angle ABC is thus less than a right- angle. And it is in segment ABC, (which is) greater than a semi-circle. And since ABCD is a quadrilateral within a circle, and for quadrilaterals within circles the (sum of the) op- posite angles is equal to two right-angles [Prop. 3.22] [angles ABC and ADC are thus equal to two right- angles], and (angle) ABC is less than a right-angle. The remaining angle ADC is thus greater than a right-angle. And it is in segment ADC, (which is) less than a semi- circle. I also say that the angle of the greater segment, (namely) that contained by the circumference ABC and the straight-line AC, is greater than a right-angle. And the angle of the lesser segment, (namely) that contained 99 ETOIXEIfiN y'. ELEMENTS BOOK 3 cpepdac; 7tepie)(o^ev7) eXdxxcov eoxlv 6piDfj<;. 'Ev xuxXtp apa f\ [iev s\ xG fjjiixuxXicp yiovia 6p^>r) eaxiv, f) 8s ev xG ^te[£ovi xuf|pcxi eXdxxwv opiS/jt;, f) 8e ev xG eXdxxovi [x^tr^iaxi] [icii^wv op-dfjc xal era f] (iev xou ^.el^ovoc; x^.r][iaxo(; [ycovla] (ici^ov [eaxlv] op-drjc;, f) 8s xou eXdxxovoc x^r^iaxoc; [yiovia] eXdxxtov op-dfje onep e8ei Set^ai. X(3'. 'Eav xuxXou scpdnxrjxai xu; euiDela, duo 8s xfjc; dcpfji; ei<; xov xuxXov 8ia)cdr) xi<; eu-dela xe^ivouaa xov xuxXov, a<; Tioiel yoviac Txpog xfj e(pan;xo^evr], faai eaovxai xalc; ev xou; evaXXdc; xou xuxXou x^ur^aai ywvlait;. KuxXou yap xou ABrA ecpanxeo'dw xu; eu-dela f\ EZ xaxd xo B ar^elov, xal duo xou B ar^dou Sir^^co T ^ cu-dela zlc, xov ABrA xuxXov xe^ivouaa auxov rj BA. Xeyto, 6xi a<; TioieT ywv[a<; f] BA ^exa xrj<; EZ ecpaTtxo^evr]<;, l'aa<; saovxai xdu; ev xolg evaXXdJ; xjiruiaoi xou xuxXou yoviau;, xouxeaxiv, oxi f) \ie\ Otto ZBA ycovia Xar] eaxl xrj sv xG BAA x|ir]jiaxi ouviaxajisv/) ywvia, f] 8e utto EBA y«v[a Tar) eaxl xfj ev xG ArB xjj.iq[iaxi auviaxa(j.evr) ycovia. "Hy^co yap duo xou B xfj EZ Ttpoc; opiSat; f] BA, xal eiXiQcp'dw era xfj<; BA 7iepicpepe(a<; xu)(6v ar)[ieTov xo T, xal sne^sux'dwaav ai AA, Ar, TB. Kal enel xuxXou xou ABrA ecpdnxexai xu; eMela f) EZ by the circumference AD[C] and the straight-line AC, is less than a right-angle. And this is immediately apparent. For since the (angle contained by) the two straight-lines BA and AC is a right-angle, the (angle) contained by the circumference ABC and the straight-line AC is thus greater than a right-angle. Again, since the (angle con- tained by) the straight-lines AC and AF is a right-angle, the (angle) contained by the circumference AZ)[C] and the straight-line CA is thus less than a right-angle. Thus, in a circle, the angle in a semi-circle is a right- angle, and that in a greater segment (is) less than a right-angle, and that in a lesser [segment] (is) greater than a right-angle. And, further, the [angle] of a seg- ment greater (than a semi-circle) [is] greater than a right- angle, and the [angle] of a segment less (than a semi- circle) is less than a right-angle. (Which is) the very thing it was required to show. Proposition 32 If some straight-line touches a circle, and some (other) straight-line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight-line) makes with the tangent will be equal to the angles in the alternate segments of the circle. A For let some straight-line EF touch the circle ABCD at the point B, and let some (other) straight-line BD have been drawn from point B into the circle ABCD, cutting it (in two) . I say that the angles BD makes with the tangent EF will be equal to the angles in the alter- nate segments of the circle. That is to say, that angle FBD is equal to the angle constructed in segment BAD, and angle EBD is equal to the angle constructed in seg- ment DCB. For let BA have been drawn from B, at right-angles to EF [Prop. 1.11]. And let the point C have been taken at random on the circumference BD. And let AD, DC, 100 ETOIXEIfiN y'. ELEMENTS BOOK 3 xaxd to B, xal dmo xfjc; acpfjc; rjxxai xfj ecpauxo^iivr] Ttpoc; opMc; f] BA, era xfjc BA apa xo xevxpov eaxl xou ABrA xuxXou. f] BA apa Bid^texoc; eaxi xou ABrA xuxXou- f] apa uno AAB ywvia ev f)|iixuxX[fc> ouaa op'dr] eaxiv. Xoiraxi apa ai bnb BAA, ABA ^tia op'dfj i'aai eiaiv. eoxl Be xal f] Otto ABZ op'Ory fj apa (mo ABZ iar) eaxi xdic; Otto BAA, ABA. xoivfj dcprprp'dco f) Otto ABA- XoiTtf] dpa f) 0ti6 ABZ ywvia Xat] eaxl xfj ev xG evaXXdJ; xjiiq^axi xou xuxXou ycovia xfj Otto BAA. xal end ev xuxXip xexpaTiXeupov eaxi xo ABrA, ai dnevavxiov aOxou ywviai Sualv op^aTc; i'aai eiaiv. eiol Be xal ai Otto ABZ, ABE Bualv opiate; i'aai- ai apa Otto ABZ, ABE xdic; Otto BAA, BEA iaai eiaiv, Sv f] Otto BAA xfj Otto ABZ eSei/i}/) far) - Xomf] dpa f) utio ABE xfj ev xG evaXXdi; xou xuxXou xjir|(jiaxi xG ArB xfj Otto ArB yovia eaxiv Tar). °Eav dpa xuxXou ecpaTtxrjxai xic; eu-dela, diro Be xfjc; acpfjc; eic; xov xuxXov 8ia)c8f) xic; eui5ela xejivouaa xov xuxXov, ac; TioieT ycoviac; Tip6<; xfj ecpauxo^ievr], iaai eaovxai xdic; ev xolc; evaXXac; xou xuxXou xjnqjiaai ywviaic;- OTtep eBei Bel^ai. and CB have been joined. And since some straight-line EF touches the circle ABCD at point B, and BA has been drawn from the point of contact, at right-angles to the tangent, the center of circle ABCD is thus on BA [Prop. 3.19]. Thus, BA is a diameter of circle ABCD. Thus, angle ADB, being in a semi-circle, is a right-angle [Prop. 3.31]. Thus, the remaining angles (of triangle ADB) BAD and ABD are equal to one right-angle [Prop. 1.32]. And ABF is also a right-angle. Thus, ABF is equal to BAD and ABD. Let ABD have been subtracted from both. Thus, the remain- ing angle DBF is equal to the angle BAD in the alternate segment of the circle. And since ABCD is a quadrilateral in a circle, (the sum of) its opposite angles is equal to two right-angles [Prop. 3.22]. And DBF and DBF is also equal to two right-angles [Prop. 1.13]. Thus, DBF and DBF is equal to BAD and BCD, of which BAD was shown (to be) equal to DBF. Thus, the remaining (angle) DBF is equal to the angle DCB in the alternate segment DCB of the circle. Thus, if some straight-line touches a circle, and some (other) straight-line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight-line) makes with the tangent will be equal to the angles in the alternate segments of the circle. (Which is) the very thing it was required to show. Ay'. 'Era xfjc; Bo-Mo/]? cu^eiac; ypd^ai xufjpt xuxXou Bexo^ie- vov yoviav ia/]v xrj Bo'deiar] ywvia etrduypdu^w. E b E "Eaxw f) Bo'Oelaa euiJeTa f] AB, f) Be BoiJeTaa yovia eO-duypajjijioc; f] Ttpoc; xG E Bel Br) era xfjc; Bo-deiarjc; eO-deiac; xfjc AB ypd(j;ai xjifj^ia xuxXou Be/ojjievov ywviav larjv xrj Ttpoc; xG r. H Bf) Ttpoc; xG r [ytovia] rjxoi 6c;e1d eaxiv f\ op'df) fj djjipXeTa- eaxw Ttpoxepov 6i;eTa, xal Gc; era xfjc; TtpGxrjc; xa- xaypacpfjc auveaxdxw Ttpoc; xfj AB eO'deia xal xG A ar)[is'\.(x> xfj Ttpoc; xG r ywvia iar) f) Otto BAA' 6<;eTa dpa eaxl xal f) utio BAA. f]xi3« xfj AA Ttpoc; opftac, f] AE, xal xexjif)ai9« f] AB 8[)(a xaxd xo Z, xal fjX'&co dno xou Z arj^ebu xfj AB Proposition 33 To draw a segment of a circle, accepting an angle equal to a given rectilinear angle, on a given straight-line. E B E Let AB be the given straight-line, and C the given rectilinear angle. So it is required to draw a segment of a circle, accepting an angle equal to C, on the given straight-line AB. So the [angle] C is surely either acute, a right-angle, or obtuse. First of all, let it be acute. And, as in the first diagram (from the left), let (angle) BAD, equal to angle C, have been constructed on the straight-line AB, at the point A (on it) [Prop. 1.23]. Thus, BAD is also acute. Let AE have been drawn, at right-angles to DA [Prop. 1.11]. 101 ETOIXEIfiN y'. ELEMENTS BOOK 3 Kpbc, op-ddc f] ZH, xal STte^eux'&o f] HB. Kal enel far) eaxiv f) AZ xfj ZB, xoivf) 8e f) ZH, 860 8f) ai AZ, ZH Suo xdic BZ, ZH faai eiaiv xal yovia f) utio AZH [yov[a] xfj utio BZH far)- pdaic dpa f) AH pdaei xfj BH lor] eaxiv. 6 dpa xevxpo ^.ev iw H 8iaaxr)[iaxi Se iu HA xuxXoc ypacpouevoc fjcei xal 8id xou B. yeypdqydo x °d eaxo 6 ABE, xal CTieCeux'do ^ EB. excel ouv dm' axpac xfjc AE Sia^iexpou duo xou A xrj AE xcpoc op'ddc eaxiv f) AA, f] A A apa ecpdicxexai xou ABE xuxXou- excel ouv xuxXou xou ABE ecpdxcxexai xu; eu'dela f) AA, xal duo xfjc xaxd xo A dcpfjc tic, xov ABE xuxXov Sifjxxai tic, cu^ela f) AB, f] apa utio AAB yovia for) eaxl xfj ev tw evaXXdc' xou xuxXou x^ir)[iaxi yovia xfj utio AEB. dXX' r\ utio AAB xfj Tipoc xo T eaxiv for)- xal f) Tipoc xo T apa yovia for] eaxl xfj utio AEB. Tkl xfjc So-deiarjc apa eui!)e(a<; xfjc AB x^if)[ia xuxXou yeypaKxai xo AEB Sexojievov yoviav xf]v utio AEB I'arjv xrj Bo'deiorj xfj Ttpoc xo T. AXXa 8f] op'df] eaxo f) Ttpoc xo E xal Seov TidXiv eaxo excl xfjc AB ypdtjjai xjifjpt xuxXou Se)(6fievov yoviav Tarjv xfj xcpoc xo T op'drj [yovia]. auveaxdxo [xcdtXiv] xrj Tcpoc xo T dp'&r] yovia I'ar) f] utio BAA, 6<; e^ei excl xfjc 8euxepac xaxa- ypacpfjc, xal xex^.rfo'do f) AB 8[)(a xaxd xo Z, xal xevxpo xo Z, 8iaaxf]^axi Se oxcoxepo xov ZA, ZB, xuxXoc yeypdcp-do 6 AEB. 'Ecpduxexai apa f] AA eu'dela xou ABE xuxXou Bid xo op'drjv eTvai xf]v Tcpoc xo A yoviav. xal for) eaxiv f] utio BAA yovia xrj ev xo AEB x^r^axi- op'df] yap tod auxrj ev f]^ixuxXio ouaa. dXXd xal f] utio BAA xfj Tipoc xo T far) eaxiv. xal f] ev to AEB apa far) lax! xfj Tipoc xo F. TcypaTixai apa TidXiv era xfjc AB x^tfjua xuxXou xo AEB 8e)(6^evov yoviav larjv xfj Tipoc xo T. AXXa 8f) f) Tip6<; xo T d^pXsTa eaxo- xal auvsaxdxo auxfj larj Tipoc xfj AB euiJeia xal xo A ar)[ie(o f) utio BAA, 6c ex £l T ^ tpttTiZ xaxaypacpfjc, xal xfj AA Tipoc op'&a.z fjx'do f) AE, xal xex^rjado TidXiv f) AB 8i^a xaxd xo Z, xal xfj AB Tip6<; op-ddc fjx'do f) ZH, xal £TieCeu)fdo rj HB. Kal ETtd TidXiv far) saxlv f) AZ xfj ZB, xal xoivf) f) ZH, 8uo 8fj ai AZ, ZH Suo xdl<; BZ, ZH faai eiaiv xal yovia f) utio AZH yovla xfj utio BZH far)- pdaig apa f) AH pdaei xfj BH far] eaxiv 6 apa xevxpo ^xsv xo H 8iaaxr)(jiaxi 8s xo HA xuxXo<; ypacp6^ie:vo<; fj^Ei xal 8id xou B. ep^sa'do 6<; 6 AEB. xal etieI xfj AE Bia^iexpo dii' dxpac Tipoc op-dac. eaxiv f) AA, f) AA apa ecpdiixexai xou AEB xuxXou. xal diio xfj<; xaxd xo A eTiacpfjc Sifjxxai f) AB- f) apa utio BAA yovia far) eaxl xfj ev xo evaXXd^ xou xuxXou x^irj^iaxi xo A0B auviaxa^ifvr) yovia. dXX' f) utio BAA yovia xfj Tip6<; xo T far) eaxiv. xal f) ev xo A0B apa x^rpaxi yovia far) eaxl xfj Tip6<; xo r. 'EtiI xfjc apa ScMar^ euiJeiac xfjc AB yeypaiixai x^fjjia xuxXou xo A0B 8exo^ievov yoviav far)v xfj Tipoc xo T- oTiep e8ei Tioifjaai. And let AB have been cut in half at F [Prop. 1.10]. And let FG have been drawn from point F, at right-angles to AB [Prop. 1.11]. And let GB have been joined. And since AF is equal to FB, and FG (is) common, the two (straight-lines) AF, FG are equal to the two (straight-lines) BF, FG (respectively). And angle AFG (is) equal to [angle] BFG. Thus, the base AG is equal to the base BG [Prop. 1.4]. Thus, the circle drawn with center G, and radius GA, will also go through B (as well as A). Let it have been drawn, and let it be (de- noted) ABE. And let EB have been joined. Therefore, since AD is at the extremity of diameter AE, (namely, point) A, at right-angles to AE, the (straight-line) AD thus touches the circle ABE [Prop. 3.16 corr.]. There- fore, since some straight-line AD touches the circle ABE, and some (other) straight-line AB has been drawn across from the point of contact A into circle ABE, angle DAB is thus equal to the angle AEB in the alternate segment of the circle [Prop. 3.32]. But, DAB is equal to C. Thus, angle C is also equal to AEB. Thus, a segment AEB of a circle, accepting the angle AEB (which is) equal to the given (angle) C, has been drawn on the given straight-line AB. And so let C be a right-angle. And let it again be necessary to draw a segment of a circle on AB, accepting an angle equal to the right- [angle] C. Let the (angle) BAD [again] have been constructed, equal to the right- angle C [Prop. 1.23], as in the second diagram (from the left). And let AB have been cut in half at F [Prop. 1.10]. And let the circle AEB have been drawn with center F, and radius either FA or FB. Thus, the straight-line AD touches the circle ABE, on account of the angle at A being a right-angle [Prop. 3.16 corr.]. And angle BAD is equal to the angle in segment AEB. For (the latter angle), being in a semi-circle, is also a right-angle [Prop. 3.31]. But, BAD is also equal to C. Thus, the (angle) in (segment) AEB is also equal to C. Thus, a segment AEB of a circle, accepting an angle equal to C, has again been drawn on AB. And so let (angle) C be obtuse. And let (angle) BAD, equal to (C), have been constructed on the straight-line AB, at the point A (on it) [Prop. 1.23], as in the third diagram (from the left) . And let AE have been drawn, at right-angles to AD [Prop. 1.11]. And let AB have again been cut in half at F [Prop. 1.10] . And let FG have been drawn, at right-angles to AB [Prop. 1.10]. And let GB have been joined. And again, since AF is equal to FB, and FG (is) common, the two (straight-lines) AF, FG are equal to the two (straight-lines) BF, FG (respectively). And an- gle AFG (is) equal to angle BFG. Thus, the base AG is 102 ETOIXEIfiN y'. ELEMENTS BOOK 3 "Eaxw 6 Bo'delc; xuxXoc; 6 ABr, f] Be Bo'deTaa ywvia eu-duYpa^cx; f] npoc, xG A- Bei 8f| &7t6 xou ABr xuxXou T^fjfia dcpeXeTv Sexo^Evov yaviav I'arjv xfj Bo'defay] ywv'a £ui}uYpd(ijj.« xfj Tipog xfi A. "Hx^w toO ABr ecpaTTTojisvr) f] EZ xaxd xo B ar^aov, xal guvsgx&xo npoc xfj ZB eu-dda xal xG 7ipo<; auxfj ar^do xG B xfj Tipoc; xG A ywvia Tar] f] 0n:6 ZBI\ 'End ouv xuxXou xou ABr dpdnxexai xic su^da f) EZ, xal duo xrjc; xaxd xo B dxacprjc Birjxxai f) Br, f] uizb ZBT apa yovia Tar] eaxi xfj sv xG BAr evaXXd?; x[if]^.axi auviaxa^iivr] yovia. dXX' f) Otto ZBr xrj npoc xG A eaxiv Xor\ % xal f] ev xG BAr apa x(if|jj.axi lot) eoxl xfj 7tp6<; xG A [ywvia] . Ako xou So-devxot; apa xuxXou xou ABr x^ifj^a dcpfjprjxai xo BAr 8e)(6^£vov ywviav I'arjv xfj BoiJeior] ywvia su'duYpd^i- [ia> xfj npoc; iu A- oTiep eSei noirjaai. t Presumably, by finding the center of ABC [Prop. 3.1], drawing a equal to the base BG [Prop. 1.4]. Thus, a circle of center G, and radius GA, being drawn, will also go through B (as well as A) . Let it go like AEB (in the third diagram from the left). And since AD is at right-angles to the di- ameter AE, at its extremity, AD thus touches circle AEB [Prop. 3.16 corr.]. And AB has been drawn across (the circle) from the point of contact A. Thus, angle BAD is equal to the angle constructed in the alternate segment AEB of the circle [Prop. 3.32]. But, angle BAD is equal to C. Thus, the angle in segment AHB is also equal to C. Thus, a segment AHB of a circle, accepting an angle equal to C, has been drawn on the given straight-line AB. (Which is) the very thing it was required to do. Proposition 34 To cut off a segment, accepting an angle equal to a given rectilinear angle, from a given circle. Let ABC be the given circle, and D the given rectilin- ear angle. So it is required to cut off a segment, accepting an angle equal to the given rectilinear angle D, from the given circle ABC. Let EF have been drawn touching ABC at point BJ And let (angle) FBC, equal to angle D, have been con- structed on the straight-line FB, at the point B on it [Prop. 1.23]. Therefore, since some straight-line EF touches the circle ABC, and BC has been drawn across (the circle) from the point of contact B, angle FBC is thus equal to the angle constructed in the alternate segment BAG [Prop. 1.32]. But, FBC is equal to D. Thus, the (angle) in the segment BAC is also equal to [angle] D. Thus, the segment BAC, accepting an angle equal to the given rectilinear angle D, has been cut off from the given circle ABC. (Which is) the very thing it was re- quired to do. ht-line between the center and point B, and then drawing EF through 103 ETOIXEIfiN y'. ELEMENTS BOOK 3 point B, at right-angles to the aforementioned straight-line [Prop. 1.11]. Xe'. 'Eav sv xuxXcp 860 su-dslai x£^iv«aiv dXXfjXac;, to utco t£>v Tfjc; (iia<; T^irpaTCOv Tcspisxojisvov op'doytoviov i'aov sail to utco t65v Tfjc; STspac; T^r)[idTWv Tispis/ojievcp op-doytoviw. 'Ev yap xuxXw to ABEA 860 sMsTai ai Ar, BA TS^vsTtoaav dXXfjXac; xaxd to E ar][Mov Xeyco, oxi to utco twv AE, Er Tcspisxojisvov 6piL>oy«viov Iaov saTi tQ utco twv AE, EB Tcspisxo|iEvw op-doycovio. El jjisv ouv ai Ar, BA 81a tou xsvTpou siaiv wots to E xevTpov slvai tou ABEA xuxXou, cpavspov, oti iacov ouaGv iSv AE, Er, AE, EB xod to utco tGv AE, Er Tcspisxojisvov op^oycoviov Iaov laxl tw utco xwv AE, EB Tcspisxo^svw op^oyMvicp. Mr] saToaav 8f) ai Ar, AB 81a tou xsvTpou, xai eiXrjcp'dw to xsvTpov tou ABrA, xai saTM to Z, xai dico tou Z era to«; Ar, AB suifteiac; xordeToi fjx'dwcrav ai ZH, Z6, xai STieCeuxtiwaav ai ZB, Zr, ZE. Kai stcsi suiMd tic; 81a tou xsvTpou f] HZ su-fMav Tiva \±r\ Sid tou xevTpou ttjv Ar Tcpoc; opMc; ts^ivsi, xai Bi^a ai)Tf|V ts[ivsi- i'ar) apa f] AH Tfj Hr. stcei ouv su'dsTa f) Ar T£T^r)Tai sic; [isv laa xaTa to H, sic 8s aviaa xaTa to E, to apa utco itov AE, Er Tcspisxo^isvov opiDoywviov ^.STa tou and Tfj<; EH TSTpaywvou 'iaov taxi tw duo Tfjc; Hr- [xoivov] Tcpoaxsia'dw to dico xrj<; HZ- to apa utco tGv AE, Er ^STa twv dico twv HE, HZ i'aov scrci toTc; dico tGv TH, HZ. dXXa toTc [Jtsv duo twv EH, HZ Iaov saTi to duo Tfjc; ZE, tou; 8s duo twv TH, HZ Iaov scrci to duo Tfjc; ZE to apa utco twv AE, Er ^.STa tou aKO Tfjc; ZE i'aov scrci tw dico Tfjc; Zr. i'ar) 8s f] ZT Tfj ZB- to apa utco twv AE, Er jxexa tou dico Tfjc; EZ Iaov saTi tw duo Tfjc; ZB. 81a Ta aika Sf] xai to utco tov AE, EB ^.STa tou dico Tfjc; ZE iaov saTi to dico Tfjc; ZB. sBsix'dr) 8s xai to utco tGv AE, Er ^iSTa tou duo Tfjc; ZE i'aov to dico Tfjc; ZB- to apa utco icov AE, Er ^iSTa tou aTco Tfjc; ZE 'iaov sot! tu utco twv AE, EB ^tSTa tou dico Tfjc; ZE. xoivov dcpfjpf]a'do to dico Tfjc; ZE- Xoitcov apa to utco twv AE, Er Tcspisxo^tsvov op-doycoviov iaov saTi to utco tCSv AE, EB Tcspisxo^svw op^oywviw. 'Eav apa sv xuxXcp su-dslai 860 TS^tvoaiv dXXfjXac;, to utco tuv Tfjc fjudg T^trpaTOv Tcspisxo^xsvov opTJoywviov iaov Proposition 35 If two straight-lines in a circle cut one another then the rectangle contained by the pieces of one is equal to the rectangle contained by the pieces of the other. For let the two straight-lines AC and BD, in the circle ABCD, cut one another at point E. I say that the rect- angle contained by AE and EC is equal to the rectangle contained by DE and EB. In fact, if AC and BD are through the center (as in the first diagram from the left), so that E is the center of circle ABCD, then (it is) clear that, AE, EC, DE, and EB being equal, the rectangle contained by AE and EC is also equal to the rectangle contained by DE and EB. So let AC and DB not be though the center (as in the second diagram from the left), and let the center of ABCD have been found [Prop. 3.1], and let it be (at) F. And let FG and FH have been drawn from F, perpen- dicular to the straight-lines AC and DB (respectively) [Prop. 1.12]. And let FB, FC, and F E have been joined. And since some straight-line, GF, through the center, cuts at right-angles some (other) straight-line, AC, not through the center, then it also cuts it in half [Prop. 3.3]. Thus, AG (is) equal to GC. Therefore, since the straight- line AC is cut equally at G, and unequally at E, the rectangle contained by AE and EC plus the square on EG is thus equal to the (square) on GC [Prop. 2.5]. Let the (square) on GF have been added [to both]. Thus, the (rectangle contained) by AE and EC plus the (sum of the squares) on GE and GF is equal to the (sum of the squares) on CG and GF. But, the (square) on FE is equal to the (sum of the squares) on EG and GF [Prop. 1.47], and the (square) on FC is equal to the (sum of the squares) on CG and GF [Prop. 1.47]. Thus, the (rectangle contained) by AE and EC plus the (square) on FE is equal to the (square) on FC. And FC (is) equal to FB. Thus, the (rectangle contained) by AE and EC plus the (square) on FE is equal to the (square) on FB. So, for the same (reasons), the (rectangle con- tained) by DE and EB plus the (square) on FE is equal 104 ETOIXEIfiN y'. ELEMENTS BOOK 3 kail xo Otto x«v xfjc sxspag x|ir]jidx«v 7iepiex°^ vt P 0?$°- ycovicp- oizep eSei Sei^ai. Xt'. 'Edv xuxXou Xr)cpa[}fj xi ar^elov exioc,, xal dm' auxoO Ttpoc; xov xuxXov TtpoaTUTtxoai 860 s&dsTou, xal f] ^tev auxov xe^vr] tov xuxXov, f\ 8s Ecpdmxrjxai, saxai to Otto oXiqc; xrj<; T£[ivo6ar)c; xal iff exxog aTraXa|jipavo|jievr]c; jisxa^u xou is a/]ue(ou xal xfjt: xupxrjc; Ttepicpepslat; I'aov to duo iff ccpa- iXTO^isvrjc; xexpayovcp. A KuxXou yap toO ABr EiXr]cp'dco ti arj^eTov exxoc; to A, xal and xou A Ttpoc; tov ABr xuxXov TtpocnuTtxexcoaav 860 eO'delai ai Ar[A], AB- xal f\ [isv ATA xeuvexo tov ABr xuxXov, f] 8s BA scpaTtxecrdw Xeyw, oil to utio xov AA, Ar ispisxo[is\iov opiJoyoviov laov kail to aTto xfj? AB xexpayovo. TL dpa [A] FA fjxoi 81a tou xevxpou eaxlv fj ou. saxo TipoTspov Sid tou xevxpou, xal eaxo to Z xevxpov tou ABr xuxXou, xal eTC^euy^co f) ZB- 6pi9r] dpa eaxlv f] utio ZBA. xal ETtel euiJJsTa #j Ar Stya xex^rjxai xaxd to Z, Ttpoaxeixai 6e aOxfj T) TA, to dpa utio xov AA, Ar \xsia tou duo iff Zr laov sail to dno xfjt; ZA. Tar) Ss f] Zr xfj ZB- xo dpa utio xov A A, Ar [isxa xou dito xfjc ZB laov eoxl xo aTto xr]<; ZA. xo 6e aTto xrjg ZA Xaa sail xd dito xov ZB, BA- xo dpa utio xov AA, Ar [ieia xou dno xfjc ZB Taov eoxl xolc; aTto xov ZB, BA. xoivov acpflprjado xo aTto xfjt; ZB- XoiTtov dpa xo utio xov AA, Ar ioov tail xo aTto xrjc; AB to the (square) on FB. And the (rectangle contained) by AE and EC plus the (square) on FE was also shown (to be) equal to the (square) on FB. Thus, the (rect- angle contained) by AE and EC plus the (square) on FE is equal to the (rectangle contained) by DE and EB plus the (square) on FE. Let the (square) on FE have been taken from both. Thus, the remaining rectangle con- tained by AE and EC is equal to the rectangle contained by DE and EB. Thus, if two straight-lines in a circle cut one another then the rectangle contained by the pieces of one is equal to the rectangle contained by the pieces of the other. (Which is) the very thing it was required to show. Proposition 36 If some point is taken outside a circle, and two straight-lines radiate from it towards the circle, and (one) of them cuts the circle, and the (other) touches (it), then the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumfer- ence, will be equal to the square on the tangent (line) . D For let some point D have been taken outside circle ABC, and let two straight-lines, DC[A] and DB, radi- ate from D towards circle ABC. And let DC A cut circle ABC, and let BD touch (it). I say that the rectangle contained by AD and DC is equal to the square on DB. [D]CA is surely either through the center, or not. Let it first of all be through the center, and let F be the cen- ter of circle ABC, and let FB have been joined. Thus, (angle) FED is a right-angle [Prop. 3.18]. And since straight-line AC is cut in half at F, let CD have been added to it. Thus, the (rectangle contained) by AD and DC plus the (square) on FC is equal to the (square) on FD [Prop. 2.6]. And FC (is) equal to FB. Thus, the (rectangle contained) by AD and DC plus the (square) on FB is equal to the (square) on FD. And the (square) on FD is equal to the (sum of the squares) on FB and BD [Prop. 1.47]. Thus, the (rectangle contained) by AD 105 ETOIXEIfiN y'. ELEMENTS BOOK 3 ecpan:xojievr)c. AXXd 8f) f\ ATA [L7] eaxw Bid xou xevxpou xou ABr xuxXou, xal eiXfjcpi^Gj to xevxpov to E, xal duo xou E eicl xf)v Ar xd-dexoc ryyj}^ f] EZ, xal eTte£eu)edioaav ai EB, Er, EA- op'&r] apa eaxlv f) Otto EBA. xal enel eu-deld xic 8id xou xevxpou f) EZ eu'deTdv xiva [if] Sid xou xevxpou xfjv Ar npoc opiDdt; xejivei, xal 8[)(a auxf]v xe^tver f\ AZ dpa xfj Zr eaxiv tar), xal end eO'deTa f] Ar xex|i/]xai 8[)(a xaxd xo Z arj^slov, icpoaxeixai 8e auxfj f] TA, xo dpa Otto ifiv AA, Ar ^texd xou duo xfjc Zr ioov eaxl x£i dno xfjc ZA. xoivov TipoaxeLai&M xo diio xfjc ZE- xo dpa Otto xov AA, Ar ^.exd xwv duo xwv TZ, ZE laov eaxl xolc duo xwv ZA, ZE. xolc 8s duo xfiv rZ, ZE laov eaxl xo duo xfjc Er- 6pi}f] yap [eaxiv] f\ Otto EZr [ycovia]- xolc 8s duo xov AZ, ZE I'aov eaxl xo duo xfjc EA- xo apa Otto x«v AA, Ar jiexd xou dno xfjc Er I'aov eaxl xw aTio xfjc; EA. for) 8e f) Er xfj EB' xo dpa Otto x65v AA, Ar jiexa xou diro xfjc EB laov eaxl xo aTro xfjc EA. tu 8s dTto xfjc EA laa eaxl xd duo xGv EB, BA- op'df] yap f\ Otto EBA ytovia- xo apa Otto xwv AA, Ar ^.exd xou dTto xfjc EB laov eaxl xolc duo xwv EB, BA. xoivov dcpypfja'do xo ctnb xfjc EB- Xoittov dpa xo Otto xwv AA, Ar laov eaxl xo duo xfjc AB. Edv dpa xuxXou Xrjcp'dfj xi or\\±eiov exxoc, xal dm' auxou Tipoc xov xuxXov TipoaTiiTixwai Suo eu'delai, xal f] ^tev auxcov xe^ivr] xov xuxXov, f\ Be ecpdux/jxai, eaxai xo Otto oX/]c xfjc xe^vouarjc xal xfjc exxoc diioXa^pavo^evr]c [icxacu xou xe ar)[ie(ou xal xfjc xupxfjc icepicpepeiac laov xw duo xfjc ecpa- Ttxo^ievr)c xexpaywvy oicep eSei Belial. AC- °Edv xuxXou Xr)(pi9fi xi a/jjielov exxoc, duo 8e xou ar][ieiou Tcpoc xov xuxXov upoaTciTcxwai 860 euiMai, xal f) y.ev auxwv xejivr) xov xuxXov, f] 8e TipoaTUTtxr), fj 8e xo Otto [xfjc] 8Xr]c xfjc xe[ivouar]c xal xfjc exxoc duoXa^pa- and DC plus the (square) on FB is equal to the (sum of the squares) on FB and BD. Let the (square) on FB have been subtracted from both. Thus, the remain- ing (rectangle contained) by AD and DC is equal to the (square) on the tangent DB. And so let DC A not be through the center of cir- cle ABC, and let the center E have been found, and let EF have been drawn from E, perpendicular to AC [Prop. 1.12]. And let EB, EC, and ED have been joined. (Angle) EBD (is) thus a right-angle [Prop. 3.18]. And since some straight-line, EF, through the center, cuts some (other) straight-line, AC, not through the center, at right-angles, it also cuts it in half [Prop. 3.3]. Thus, AF is equal to FC. And since the straight-line AC is cut in half at point F, let CD have been added to it. Thus, the (rectangle contained) by AD and DC plus the (square) on FC is equal to the (square) on FD [Prop. 2.6]. Let the (square) on FE have been added to both. Thus, the (rectangle contained) by AD and DC plus the (sum of the squares) on CF and FE is equal to the (sum of the squares) on FD and FE. But the (square) on EC is equal to the (sum of the squares) on CF and FE. For [angle] EFC [is] a right-angle [Prop. 1.47]. And the (square) on ED is equal to the (sum of the squares) on DF and FE [Prop. 1.47]. Thus, the (rectangle contained) by AD and DC plus the (square) on EC is equal to the (square) on ED. And EC (is) equal to EB. Thus, the (rectan- gle contained) by AD and DC plus the (square) on EB is equal to the (square) on ED. And the (sum of the squares) on EB and BD is equal to the (square) on ED. For EBD (is) a right-angle [Prop. 1.47]. Thus, the (rect- angle contained) by AD and DC plus the (square) on EB is equal to the (sum of the squares) on EB and BD. Let the (square) on EB have been subtracted from both. Thus, the remaining (rectangle contained) by AD and DC is equal to the (square) on BD. Thus, if some point is taken outside a circle, and two straight-lines radiate from it towards the circle, and (one) of them cuts the circle, and (the other) touches (it), then the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumfer- ence, will be equal to the square on the tangent (line). (Which is) the very thing it was required to show. Proposition 37 If some point is taken outside a circle, and two straight-lines radiate from the point towards the circle, and one of them cuts the circle, and the (other) meets (it), and the (rectangle contained) by the whole (straight- 106 ETOIXEIfiN y'. ELEMENTS BOOK 3 \onevr)c, [isxa^u xoO te arpdou xal xfjc xupxfjc; Tiepicpepdac; I'aov xfi duo xfjc; upoaumxouarjc;, f\ TtpocnuTtxouaa ecpdcjjexai xoO xuxXou. KuxXou yap xoO ABr riXfjcp'StL) xi ar^eiov exxo<; xo A, xal ooto xoO A 7ip6<; xov ABr xuxXov TtpocnuTixexcoaav 860 eu-deiai ai ArA, AB, xal rj \iev ATA xe^ivexw xov xuxXov, f] 8e AB Tipoaumxexw, eaxco 8e xo Otto xCSv AA, Ar laov tu dira xfj? AB. Xeyto, oxi r) AB eqxbtxexai xou ABr xuxXou. "H)cdw yap xou ABr ecpajixo^ievr) f] AE, xal dAr|cp'dco xo xevxpov xoO ABr xuxXou, xal eaxw xo Z, xal £Tie^£U)cdwaav ai ZE, ZB, ZA. f) dpa utto ZEA op-dr] eaxiv. xal end f) AE scpotTixexai xou ABr xuxXou, xejivei 8s f) ArA, xo dpa Otto xfiv A A, Ar I'oov soxl x« aTio xfjc; AE. rjv 8s xal xo Otto xfiv A A, Ar taov xG dno xfjc; AB' xo dpa duo xfjc; AE I'oov eoxl xG duo xfjc; AB' Tar) dpa f] AE xfj AB. eoxl 8e xal f) ZE xfj ZB lor\- 860 8f) ai AE, EZ 860 xau; AB, BZ i'oai eiaiv xal pdaig auxaiv xoivf) f) ZA- ywvia dpa f] Otto AEZ ywvia xfj utto ABZ saxiv for), op-df] Ss f) Otto AEZ- 6pi9r) dpa xal f) utio ABZ. xai eoxiv rj ZB sxpaXXo^evr) Sidjiexpoc;- f) 8e xfj 8ia[iexpcp xou xuxXou Ttpoc; opiJac; an' dxpac; dyo^isvr) scpaTtxexai xou xuxXou- f] AB dpa ecpaTtxexai xou ABr xuxXou. ojioimc; 8/) Ssix^iqoexai, xdv xo xsvxpov Era xfjc; Ar xuy)(dvr]. °Edv dpa xuxXou X/]cpi9fj xi a/j^islov sxxog, arco 8s xou ar][ieio\j npbc, xov xuxXov TipooTiiTixwai 860 euiSdai, xal f] y.sv auxfiv xspivr) xov xuxXov, f) 8e TipooniTixr), rj 8e xo utio oXrjc; xfjc xe[ivouar)<; xal xfjc exxoc; dTioXajipavojisvrjc; ^exac;u xou xs af][ieioyj xal xfjc; xupxfjc; Tispicpspdac; Taov xcp aTto xfjc; TtpoaTUTtxouarjc;, f] TtpocnuTtxouaa ecpd^sxai xou xuxXou- oTiep sSei Sdc;ai. line) cutting (the circle), and the (part of it) cut off out- side (the circle), between the point and the convex cir- cumference, is equal to the (square) on the (straight-line) meeting (the circle), then the (straight-line) meeting (the circle) will touch the circle. D For let some point D have been taken outside circle ABC, and let two straight-lines, DC A and DB, radiate from D towards circle ABC, and let DC A cut the circle, and let DB meet (the circle) . And let the (rectangle con- tained) by AD and DC be equal to the (square) on DB. I say that DB touches circle ABC. For let DE have been drawn touching ABC [Prop. 3.17], and let the center of the circle ABC have been found, and let it be (at) F. And let FE, FB, and FD have been joined. (Angle) FED is thus a right-angle [Prop. 3.18]. And since DE touches circle ABC, and DC A cuts (it), the (rectangle contained) by AD and DC is thus equal to the (square) on DE [Prop. 3.36]. And the (rectangle contained) by AD and DC was also equal to the (square) on DB. Thus, the (square) on DE is equal to the (square) on DB. Thus, DE (is) equal to DB. And FE is also equal to FB. So the two (straight-lines) DE, EF are equal to the two (straight-lines) DB, BF (re- spectively) . And their base, FD, is common. Thus, angle DEF is equal to angle DBF [Prop. 1.8]. And DEF (is) a right-angle. Thus, DBF (is) also a right-angle. And FB produced is a diameter, And a (straight-line) drawn at right-angles to a diameter of a circle, at its extremity, touches the circle [Prop. 3.16 corr.]. Thus, DB touches circle ABC. Similarly, (the same thing) can be shown, even if the center happens to be on AC. Thus, if some point is taken outside a circle, and two straight-lines radiate from the point towards the circle, and one of them cuts the circle, and the (other) meets (it), and the (rectangle contained) by the whole (straight- line) cutting (the circle), and the (part of it) cut off out- side (the circle), between the point and the convex cir- cumference, is equal to the (square) on the (straight-line) meeting (the circle), then the (straight-line) meeting (the circle) will touch the circle. (Which is) the very thing it 107 ETOIXEiQN y\ ELEMENTS BOOK 3 was required to show. 108 ELEMENTS BOOK 4 Construction of Rectilinear Figures In and Around Circles 109 ETOIXEIfiN 5'. ELEMENTS BOOK 4 "Opoi. a'. Sx^K a £ui}uypa|ji[iov etc; a/f^oc eMuypa^ov eyypdcp- ea-&ai Xeyexai, oxav exdaxr] x£>v xou eyypacpo^ievou axr^ax- oc, yiovifiv exdaxrjc; TiXeupac; tou, etc; o eyypdcpexai, dTtxrjxai. P'. S)(fi(jia 8e 6|jioi«c; Ttepl oxrjjia Ttepiypdcpecrdai Xeyexai, oxav exdaxr] TiXeupa xoO Tiepiypacpojievou exdaxrjc; ywviac tou, tie pi 8 Tiepiypdcpexai, cbixrjxai. y'. E/rjiia eMuypajjijiov etc; xuxXov eyypdcpecrdaa Xeyexai, oxav exdaxr) ywvia xou eyypacpojievou aTixrjxai xrjc; xou xuxXou Tiepicpepeiac;. 8'. S/'^l jla Se eu-duypa^iov uepl xuxXov Ttepiypdcpe- a-Qai Xeyexai, oxav exdaxr) TiXeupa xou Tiepiypacpojievou ecpaTix/jxai xrjc; xou xuxXou Ttepicpepeiac;. e'. KuxXog 8e etc; a^/j^a 6\±oiu>q eyypdcpeo'dai Xeyexai, oxav f) xou xuxXou Ttepicpepeia exdaxrjc; TiXeupac; xou, etc; o eyypdcpexai, aTix/jxai. <?'. KuxXoc 8e Tiepl axrjjia Tiepiypdcpecdai Xeyexai, oxav f) xou xuxXou Tiepicpepeia exdaxrjc; ywviac; xou, uepl o Tie- piypdcpexai, aTixrjxai. Eu-deTa etc; xuxXov evapjio^ecrdai Xeyexai, oxav xa nepaxa auxrjg era xrjc; Tiepicpepeiac; fj xou xuxXou. a . Etc; xov 8oi3evxa xuxXov xfj Soi5eiarj eu-deia \lt\ [isiZovi ouar] xrjc; xou xuxXou Biajiexpou larjv eui&elav evapjioaai. A "Eaxw 6 Bo-delc; xuxXoc 6 ABr, f) 8e SoTJelaa euiSeTa ^tf) (jiei^wv xrjc xou xuxXou 8ia[iexpou f] A. 8el 8f) etc xov ABr xuxXov xfj A eu-deia iar]v eu-delav evap^toaai. "H)cda) xou ABr xuxXou 8idpiexpo<; f] Br. et uev ouv tar] eaxlv f] Br xfj A, yeyovoc; av eir) xo eTiixax^ev evrpuoaxai Definitions 1. A rectilinear figure is said to be inscribed in a(nother) rectilinear figure when the respective angles of the inscribed figure touch the respective sides of the (figure) in which it is inscribed. 2. And, similarly, a (rectilinear) figure is said to be cir- cumscribed about a(nother rectilinear) figure when the respective sides of the circumscribed (figure) touch the respective angles of the (figure) about which it is circum- scribed. 3. A rectilinear figure is said to be inscribed in a cir- cle when each angle of the inscribed (figure) touches the circumference of the circle. 4. And a rectilinear figure is said to be circumscribed about a circle when each side of the circumscribed (fig- ure) touches the circumference of the circle. 5. And, similarly, a circle is said to be inscribed in a (rectilinear) figure when the circumference of the circle touches each side of the (figure) in which it is inscribed. 6. And a circle is said to be circumscribed about a rectilinear (figure) when the circumference of the circle touches each angle of the (figure) about which it is cir- cumscribed. 7. A straight-line is said to be inserted into a circle when its extemities are on the circumference of the circle. Proposition 1 To insert a straight-line equal to a given straight-line into a circle, (the latter straight-line) not being greater than the diameter of the circle. D Let ABC be the given circle, and D the given straight- line (which is) not greater than the diameter of the cir- cle. So it is required to insert a straight-line, equal to the straight-line D, into the circle ABC. Let a diameter BC of circle ABC have been drawn.t 110 ETOIXEIfiN 5'. ELEMENTS BOOK 4 yap zlc, xov ABr xuxXov xfj A sinJsia Tor) f] BY. zl Bs (isi^cov soxlv f) Br xfjc A, xslo'dw xfj A lor] f] TE, xod xsvxpw xto T 8iaoxf|[iaxi 8s xc5 TE xuxXoc; ysypdcpiSio 6 EAZ, xod £Tie^£U)cdw f) TA. 'End ouv to T ar^slov xsvxpov soxl xoO EAZ xuxXou, far) soxlv f] TA xfj TE. dXXd xfj A f] TE soxiv for) - xal f) A apa xfj TA soxiv Tor). Etc; apa xov Bo'dsvxa xuxXov xov ABr xfj Bo-dslor] sui5sia xfj A tor] Evrpuooxai f] TA- oTisp sSsi Ttoifjoai. Therefore, if £?C is equal to D then that (which) was prescribed has taken place. For the (straight-line) BC, equal to the straight-line D, has been inserted into the circle ABC. And if BC is greater than D then let CE be made equal to D [Prop. 1.3], and let the circle EAF have been drawn with center C and radius CE. And let CA have been joined. Therefore, since the point C is the center of circle EAF, CA is equal to CE. But, CE is equal to D. Thus, D is also equal to CA. Thus, CA, equal to the given straight-line D, has been inserted into the given circle ABC. (Which is) the very thing it was required to do. Presumably, by finding the center of the circle [Prop. 3.1], and then drawing a line through it. P'- Etc xov 8oi9svxa xuxXov xw Scdevxi xpiycovcp Eooycoviov xptyovov syypd(|>ai. 'Eoxco 6 So'delc; xuxXoc; 6 ABr, xo 8s 8oif)sv xpiywvov xo AEZ- 8sT 5f] zlc, xov ABr xuxXov xw AEZ xpiywvcp looyoviov xplyovov eyypd^ai. "H)cda) xou ABr xuxXou scpaTixo^iivr] f] H0 xaxd xo A, xal ouvsoxdxco npbc, xfj A9 sO'dsia xal xw Ttpoc; auxfj or^slo tu A xfj utio AEZ ywvia Tar) f] utio 0Ar, Ttpoc 8s xfj AH sui5sia xal xa> Tipoc auxfj or)u.s(cp xfi A xfj utio AZE [ycovia] Tor] f] utio HAB, xal etis^sux'&co f) BE 'End ouv xuxXou xou ABT scpdTtxsxal xic su-dsla f) A0, xal duo xrjc xaxd xo A STiacprjc sic; xov xuxXov Birjxxai sO'dsia f) Ar, f) apa utio 9Ar Tor) soxl xfj sv x£5 svaXXdi; xou xuxXou x^f][iaxi ywvia xfj utio ABr. dXX' rj utio 0Ar xfj utio AEZ soxiv lor]' xal r) utio ABr apa ywvia xfj utio AEZ soxiv Tor). Bid xd auxa 8f) xal f] utio ArB xfj utio AZE soxiv lor]- xal XoiTif] apa f) utio BAr Xomfj xfj utio EAZ soxiv lor) [looywviov apa eoxl xo ABr xpiywvov ifi AEZ xpiywvw, xal syysypanxai sic tov ABT xuxXov]. Etc xov Bo'dsvxa apa xuxXov ifi 8o , f)svxi xpiycovcp looyoviov xpiycovov syysypaiixai- OTisp sBsi Tioifjaai. Proposition 2 To inscribe a triangle, equiangular with a given trian- gle, in a given circle. ^\ E Let ABC be the given circle, and DEF the given tri- angle. So it is required to inscribe a triangle, equiangular with triangle DEF, in circle ABC. Let GH have been drawn touching circle ABC at A.^ And let (angle) HAC, equal to angle DEF, have been constructed on the straight-line AH at the point A on it, and (angle) GAB, equal to [angle] DFE, on the straight- line AG at the point A on it [Prop. 1.23]. And let BC have been joined. Therefore, since some straight-line AH touches the circle ABC, and the straight-line AC has been drawn across (the circle) from the point of contact A, (angle) HAC is thus equal to the angle ABC in the alternate segment of the circle [Prop. 3.32]. But, HAC is equal to DEF. Thus, angle ABC is also equal to DEF. So, for the same (reasons), ACB is also equal to DFE. Thus, the re- maining (angle) BAG is equal to the remaining (angle) EDF [Prop. 1.32]. [Thus, triangle ABC is equiangu- lar with triangle DEF, and has been inscribed in circle 111 ETOIXEIfiN 5'. ELEMENTS BOOK 4 t See the footnote to Prop. 3.34. y'- Ilepl xov Bo-devxa xuxXov iu Bo'devxi xpiyovo iaoyoviov xpiywvov Tiepiypd^ai. a r n "EaTO 6 Scdslc xuxXo<; 6 ABr, to Se So-fJev xpiycovov to AEZ- 8eT 5r| Tiepi xov ABr xuxXov iw AEZ xpiycovcp iaoycoviov xpiyovov Tiepiypdtjiai.. 'ExpEpX^a-dw f] EZ eq>' exdxepa xd [ispf] xaxd xd H, 6 ar^eia, xal eiXr]q)T&to tou ABr xuxXou xevxpov to K, xod Bi^jcda), (be; etu)(£v, euiDeTa f\ KB, xai auvsaxdxo Tipoc; xfj KB EU'deia xal xG Tipoc; auxfj ar^dw iu K xfj fjisv utio AEH yovia Xar] r) utio BKA, xfj 8e utio AZ9 iar) f] utio BKr, xal 8id xfiv A, B, T ar^eiwv f))fdwoav scpaTixojievai xou ABr xuxXou ai AAM, MBN, NEA Kal euel eqxbixovxai xou ABr xuxXou ai AM, MN, NA xaxd xa A, B, r ar^eia, duo 8e xou K xsvxpou era xd A, B, r ar^eia ETieCsuy^evai elolv ai KA, KB, Kr, bp-Qai apa eialv ai Tipoc; xolc; A, B, T ar^eioiz ywviai. xal etiei xou AMBK xexpauXeupou ai xeaaapec; ycoviai xexpaaiv op-ddic; law. eiaiv, £7i£i8r]Tiep xal tic, Suo xpiywva Siaipdxai xo AMBK, xai siaiv opiDal ai utio KAM, KBM ywviai, Xomai apa ai utio AKB, AMB 8uaiv opiate; i'aai eiaiv. riai Be xai ai utio AEH, AEZ Suaiv opiJau; i'aai- ai apa utio AKB, AMB xdic; utio AEH, AEZ Taai eiaiv, £>v f] utio AKB xfj utio AEH eaxiv lot]- XoiTif) apa r] utio AMB XoiTifj xfj utio AEZ eaxiv Xar). 6^ioi«<; 8r) Beix^^asxai, oxi xai f] utio ANB xfj utio AZE eaxiv Xar}' xai XoiTif) apa f) utio MAN [Xomfj] xfj utio EAZ eaxiv Xar). iaoyoviov apa eaxi xo AMN xpiywvov x£> AEZ xpiyovw- xai TiepiyeypaTixai Tiepi xov ABr xuxXov. nepl xov Bo'devxa apa xuxXov x£> 8oif)evxi xpiywvw iaoyoviov xpiyovov TiepiyeypaTixar oiiep e8ei Tioifjaai. ABC]. Thus, a triangle, equiangular with the given triangle, has been inscribed in the given circle. (Which is) the very thing it was required to do. Proposition 3 To circumscribe a triangle, equiangular with a given triangle, about a given circle. H 7 D L C N Let ABC be the given circle, and DEF the given tri- angle. So it is required to circumscribe a triangle, equian- gular with triangle DEF, about circle ABC. Let EF have been produced in each direction to points G and H. And let the center K of circle ABC have been found [Prop. 3.1]. And let the straight-line KB have been drawn, at random, across {ABC). And let (angle) BKA, equal to angle DEC, have been con- structed on the straight-line KB at the point K on it, and (angle) BKC, equal to DFH [Prop. 1.23]. And let the (straight-lines) LAM, MBN, and NCL have been drawn through the points A, B, and C (respectively), touching the circle ABC'J And since LM, MN, and NL touch circle ABC at points A, B, and C (respectively), and KA, KB, and KC are joined from the center K to points A, B, and C (respectively), the angles at points A, B, and C are thus right-angles [Prop. 3.18]. And since the (sum of the) four angles of quadrilateral AMBK is equal to four right- angles, inasmuch as AMBK (can) also (be) divided into two triangles [Prop. 1.32], and angles KAM and KBM are (both) right-angles, the (sum of the) remaining (an- gles), AKB and AMB, is thus equal to two right-angles. And DEC and DEF is also equal to two right-angles [Prop. 1.13]. Thus, AKB and AMB is equal to DEC and DEF, of which AKB is equal to DEC. Thus, the re- mainder AMB is equal to the remainder DEF. So, sim- ilarly, it can be shown that LNB is also equal to DFE. Thus, the remaining (angle) MLN is also equal to the 112 ETOIXEIfiN 5'. ELEMENTS BOOK 4 t See the footnote to Prop. 3.34. 5'. Etc to 6otl>ev Tplycovov xuxXov eryypd^ai. A T5gto> to SoiSev Tpiywvov to ABE 8a 5r] el? to ABr Tpiywvov xuxXov eyypdijiai. TeT^nqcrdMaav od Otto ABr, ArB ycoviai 8i)(a Tdig BA, TA su-vMaig, xod aujipaXXsTwaav dXXr|Xaig xaTa to A o/jjisTov, xal fj/iJioaav duo tou A km Tag AB, Br, EA sMelag xdiJeToi al AE, AZ, AH. Kod inz\ lot) eotIv f\ uno ABA ytovla T/j utto TBA, eaTi 8e xai op'dr) i\ uno BEA op^fj t/j utto BZA lot], 86o 8r] Tpiycovd eoTi Td EBA, ZBA Tag 80o ycovlag Talc; Sual ycoviaig laag e)( 0VTa xal ( i ' av ^Xeupdv [iia TiXsupa Taiqv tt]v OnoTEivouaav Otto [ilav twv lacov ywviaiv xoivfjv auTfiv tt]v BA- xal Tag XoiTidg apa TcXsupag Tdig XoiTtaTg TcXsupdig laag sg'ouaiv Tor) apa f] AE Tfj AZ. Bid Ta ai)Ta 8f) xal f) AH tt) AZ eoTiv Tor), al Tpelg apa euifeTai al AE, AZ, AH I'oai dXXr]Xaig eloiv 6 apa xsvTpG tG A xal 8iaoTr]jj.aTi evl tGv E, Z, H xuxXog ypacpojisvog r^gei xal 8id tcov XoitcSSv a/][id«v xal scpdcJjSTai t«v AB, Br, TA euifteiwv 8id to op-ddg dvai Tag Tcpog Tolg E, Z, H ar^eion; yioviag. el yap T£[iei auTag, scxuai f\ Tfj Bia^STpcp tou xuxXou Ttpog opiDdg an' axpag dyojievr) evTog TUTCTOuaa tou xuxXou- onep aTO- tiov £8el)cdr)- oux apa 6 xevTpip t£5 A SiarjTrj^aTi Se evl tGv E, Z, H ypacpojisvog xuxXog TE^iel Tag AB, Br, TA eu^dag' £(pd(|j£Tai apa auTfiv, xal scnai 6 xuxXog eyyeypajijisvog rig to ABr Tplytovov. syyeypdcp-dw tbg 6 ZHE. Elg apa to 8o$ev Tpiywvov to ABr xuxXog eyyeypaKTai 6 EZH- oTiep eBsi noifjaaL. [remaining] (angle) EDF [Prop. 1.32]. Thus, triangle LMN is equiangular with triangle DEF. And it has been drawn around circle ABC. Thus, a triangle, equiangular with the given triangle, has been circumscribed about the given circle. (Which is) the very thing it was required to do. Proposition 4 To inscribe a circle in a given triangle. A Let ABC be the given triangle. So it is required to inscribe a circle in triangle ABC. Let the angles ABC and ACB have been cut in half by the straight-lines BD and CD (respectively) [Prop. 1.9], and let them meet one another at point D, and let DE, DF, and DC have been drawn from point D, perpendic- ular to the straight-lines AB, BC, and CA (respectively) [Prop. 1.12]. And since angle ABD is equal to CBD, and the right- angle BED is also equal to the right-angle BFD, EBD and FBD are thus two triangles having two angles equal to two angles, and one side equal to one side — the (one) subtending one of the equal angles (which is) common to the (triangles) — (namely), BD. Thus, they will also have the remaining sides equal to the (corresponding) remain- ing sides [Prop. 1.26]. Thus, DE (is) equal to DF. So, for the same (reasons), DC is also equal to DF. Thus, the three straight-lines DE, DF, and DC are equal to one another. Thus, the circle drawn with center D, and radius one of E, F, or Gj will also go through the re- maining points, and will touch the straight-lines AB, BC, and CA, on account of the angles at E, F, and G being right-angles. For if it cuts (one of) them then it will be a (straight-line) drawn at right-angles to a diameter of the circle, from its extremity, falling inside the circle. The very thing was shown (to be) absurd [Prop. 3.16]. Thus, the circle drawn with center D, and radius one of E, F, 113 ETOIXEIfiN 5'. ELEMENTS BOOK 4 or G, does not cut the straight-lines AB, BC, and CA. Thus, it will touch them and will be the circle inscribed in triangle ABC. Let it have been (so) inscribed, like FGE (in the figure). Thus, the circle EFG has been inscribed in the given triangle ABC. (Which is) the very thing it was required to do. t Here, and in the following propositions, it is understood that the radius is actually one of DE, DF, or DG. e . Ilepi to Bo'dev Tpiycovov xuxXov 7iepiypd<j>ai. "Eaxw to SoTJev xpiywvov to ABE 8a Se rcspl to Bo'dev Tpiycovov to ABT xuxXov 7tepiypd(];ai. TeT^iiqo'Owoav ai AB, Ar euifteTai Si)(a xaTa Ta A, E ar^ela, xai arco tcov A, E arjueitov xau; AB, AT Tipoc. 6pi9dc; f])Cdwoav ai AZ, EZ- aupmeaouvTai 5r) f]Toi evTo<; toO ABr Tpiycovou f] em Trj<; Br eO'deiac. fj cxtoc. Trjc. Br. Sujj.7ii7iT£TCdaav upoTEpov Ivtoc. xaTa to Z, xai £Tie^£U)cd- coaav ai ZB, ZT, ZA. xai end Tar) eaTiv f) AA Tfj AB, xoivr) 8e xai Ttpoc. opiSdc. f] AZ, pdai<; dpa f] AZ pdaei Tfj ZB eaTiv lor\. b[io'ux>z 8t) Bei^o^iev, oti xai f] TZ T/j AZ ecrciv tar]- 6aie xai f] ZB Tfj ZT eaTiv Tar]- ai TpeX? dpa ai ZA, ZB, ZT laai dXXf|Xaic. eiaiv. 6 apa xevTpco tw Z 8iaaTf|[iaTi 8e evi twv A, B, r xvxkoc, ypacpojievoc; fj^ei xai 8ia tcov XoitcSSv ar^eicov, xai eaTai 7iepiyeypa[i(jievoc. 6 xuxXoc; Tiepi to ABr Tpiywvov. Tiepiyeypdcp-dco cbc. 6 ABE AXXa 8r) ai AZ, EZ aupmmTSTwaav iiu Tfjc. BT eu-Mac. xaTa to Z, cbc; ex £l ^ BeuTepac. xaTaypacpfjc,, xai CTieCeux'dM ^ AZ. 6[ioicoc 8f] 8eic;o[iev, oti to Z aruielov xevTpov eaTi tou Ttepi to ABT Tpiycovov Ttepiypacpo^ievou xuxXou. AXXa 8r] ai AZ, EZ aujjiTimTSTwaav cxtoc. toO ABT Tpiycivou xaTa to Z TtdXiv, cbc. l/ei em Trjc. TpiTr]C xaTa- ypacprjc,, xai eTte£eu)cd«aav ai AZ, BZ, TZ. xai eizsl TtdXiv lot] eaTiv f] AA Tfj AB, xoivf) Be xai Tipoc, op'dac, f] AZ, p&au; dpa f] AZ pdaei xfj BZ eaTiv i'ar). ojioimc, 8f] Sei^o^iev, oti xai f] TZ Tfj AZ eaTiv i'ar)- waTe xai f] BZ Tfj ZT eaTiv for) - 6 dpa [rcdXiv] xevTpw t£> Z SiaaT^aTi 8e evi itov ZA, ZB, ZT xuxXoc ypacpo^ievoc ffesi xai 8id t£Sv Xoitiwv ar^eiwv, xai eaTai Ttepiyeypa[i^evoc Ttepi to ABT Tpiycovov. Ilepl to Bo-dev apa Tpiywvov xuxXo^ uepiyeypaKTar OTiep e8ei Koifjaai. Proposition 5 To circumscribe a circle about a given triangle. Let ABC be the given triangle. So it is required to circumscribe a circle about the given triangle ABC. Let the straight-lines AB and AC have been cut in half at points D and E (respectively) [Prop. 1.10]. And let DF and EF have been drawn from points D and E, at right-angles to AB and AC (respectively) [Prop. 1.11]. So (DF and EF) will surely either meet inside triangle ABC, on the straight-line BC, or beyond BC. Let them, first of all, meet inside (triangle ABC) at (point) F, and let FB, FC, and FA have been joined. And since AD is equal to DB, and DF is common and at right-angles, the base AF is thus equal to the base FB [Prop. 1.4]. So, similarly, we can show that CF is also equal to AF. So that FB is also equal to FC. Thus, the three (straight-lines) FA, FB, and FC are equal to one another. Thus, the circle drawn with center F, and radius one of A, B, or C, will also go through the remaining points. And the circle will have been circumscribed about triangle ABC. Let it have been (so) circumscribed, like ABC (in the first diagram from the left) . And so, let DF and EF meet on the straight-line BC at (point) F, like in the second diagram (from the left). And let AF have been joined. So, similarly, we can show that point F is the center of the circle circumscribed about triangle ABC. And so, let DF and EF meet outside triangle ABC, again at (point) F, like in the third diagram (from the left). And let AF, BF, and CF have been joined. And, again, since AD is equal to DB, and DF is common and at right-angles, the base AF is thus equal to the base BF [Prop. 1.4]. So, similarly, we can show that CF is also equal to AF. So that BF is also equal to FC. Thus, 114 ETOIXEIfiN 5'. ELEMENTS BOOK 4 [again] the circle drawn with center F, and radius one of FA, FB, and FC, will also go through the remaining points. And it will have been circumscribed about trian- gle ABC. Thus, a circle has been circumscribed about the given triangle. (Which is) the very thing it was required to do. Proposition 6 Etc; xov Boi9svxa xuxXov xexpdywvov eyypdijiai. A To inscribe a square in a given circle. A "Eaxw f] Bo'vteu; xuxXoc; 6 ABrA- Bel 8r) zlc. xov ABrA xuxXov xsxpdyiovov eyypdtjiai. "H)(Tf)cL)acxv xou ABrA xuxXou Buo Bidjiexpoi npbc, opiSdc; dXXfjXaic; ad Ar, BA, xod eiteCeux'dwaav al AB, Br, TA, AA. Kal em\lar\ eaxlv f] BE xfj EA- xevxpov yap to E- xoivrj Be xal Tipoc; 6pi9dc; f] EA, f3dai<; dpa f) AB pdoei xrj AA Tar) eaxiv. Sid xd auxd 5r] xal exaxepa x£5v Br, TA exaxepa xfiv AB, AA for) eoxiv iooTiXeupov dpa eoxl xo ABrA xexpaTiXsrupov. Xsy« 8r], oxi xal 6p$oywviov. inel yap f] BA su-dela Sidjisxpoc; eoxi xou ABrA xuxXou, /jjiixuxXiov dpa eaxl xo BAA- 6pi3r) dpa #j utto BAA ycovia. Sid xa auxd Br] xal exdaxr) xGv utio ABr, BrA, TAA op-Qr] soxiv 6p$oycoviov dpa eaxl xo ABTA xexpdnXeupov. eBsix'dif) Be xal EaonXsupov xsxpdytovov dpa saxiv. xal syysypanxai eic, xov ABrA xuxXov. Eic; dpa xov 8oi5evxa xuxXov xsxpdywvov syyeyparcxai xo ABrA- ojiep eBsi noifjaai. Let ABCD be the given circle. So it is required to inscribe a square in circle ABCD. Let two diameters of circle ABCD, AC and BD, have been drawn at right-angles to one another, t And let AB, BC, CD, and DA have been joined. And since BE is equal to ED, for E (is) the center (of the circle), and EA is common and at right-angles, the base AB is thus equal to the base AD [Prop. 1.4]. So, for the same (reasons), each of BC and CD is equal to each of AB and AD. Thus, the quadrilateral ABCD is equilateral. So I say that (it is) also right-angled. For since the straight-line BD is a diameter of circle ABCD, BAD is thus a semi-circle. Thus, angle BAD (is) a right- angle [Prop. 3.31]. So, for the same (reasons), (angles) ABC, BCD, and CD A are also each right-angles. Thus, the quadrilateral ABCD is right-angled. And it was also shown (to be) equilateral. Thus, it is a square [Def. 1.22]. And it has been inscribed in circle ABCD. Thus, the square ABCD has been inscribed in the given circle. (Which is) the very thing it was required to do. t Presumably, by finding the center of the circle [Prop. 3.1], drawing a line through it, and then drawing a second line through it, at right-angles to the first [Prop. 1.11]. Ilepl xov So-devxa xuxXov xexpdywvov uepiypatjiai. Proposition 7 To circumscribe a square about a given circle. 115 ETOIXEIfiN 5'. ELEMENTS BOOK 4 'Eaxw 6 So-del; xuxXo<; 6 ABrA- Bel 8rj Tiepi xov ABrA xuxXov xexpdycovov Ttepiypdijiai. H A Z B @ r k "HyT&coaav xou ABrA xuxXou 8uo Bid^expoi Tipoc; 6pi9dc; dXXf|Xai<; ai Ar, BA, xai Bid iSv A, B, T, A arjjjidcov rj/iiko- aav ecpairxo^ievai xoO ABrA xuxXou ai ZH, H6, 6K, KZ. 'Etc! ouv ecpdrcxexai rj ZH xou ABrA xuxXou, duo 8s xou E xevxpou era. xr]v xaxd xo A S7ioccpf)v eiteCeuxxai f] EA, ai apa npoc, xw A ywviai op-dai eiaiv. Bid xa auxa 8rj xai ai irpoi; xoT? B, T, A ar^eiou; ytoviai opiJai eiaiv. xai £7te! op-Qi] eaxiv f] (mo AEB ycovia, eaxi Be op'dr] xal rj utio EBH, TiapdXX/jXoi; apa eaxiv f] H9 xfj Ar. 8id xd auxa 8f) xai r) Ar xfj ZK eaxi 7tapdXXr]Xo<;. uoxe xai f) H9 xfj ZK eaxi TtapdXXr)Xo<;. ojioiwc; Bf] Bei^o^iev, oxi xal exaxepa xwv HZ, 0K xfj BEA eaxi TtapdXXr)Xo<;. 7tapaXXr]X6ypa[ipc apa eaxi xa HK, Hr, AK, ZB, BK- i'ar) apa eaxiv f) [lev HZ xfj 0K, f) Be H9 xrj ZK. xai ercel iarj eaxiv f) Ar xfj BA, dXXa xai f) jiev Ar exaxepa xGv H6, ZK, r] Be BA exaxepa xfiv HZ, 9K eaxiv for] [xai exaxepa apa xwv H0, ZK exaxepa x£>v HZ, 0K eaxiv for]], iaouXeupov apa eaxi xo ZH0K xexpduXeupov. Xeyco Br), oxi xai op-doywviov. enel yap TtapaXXr)X6ypa^6v eaxi xo HBEA, xai eaxiv op'df] f) utio AEB, op-df) apa xal f] utio AHB. 6[io[w<; Br] Bei^oiJiev, oxi xai ai npbz toiz O, K, Z ywviai op'dai eiaiv. 6p , doy«viov apa eaxi xo ZH6K. eBeix^r] Be xai iaoTiXeupov xexpdyovov apa eaxiv. xal TtepiyeypaTtxai uepl xov ABrA xuxXov. nepl xov So-devxa apa xuxXov xexpdyovov TtepiyeypaTtxai- OTiep e8ei Ttoirjaai. Let ABCD be the given circle. So it is required to circumscribe a square about circle ABCD. G A B D H C K Let two diameters of circle ABCD, AC and BD, have been drawn at right-angles to one another. ^ And let FG, GH, HK, and KF have been drawn through points A, B, C, and D (respectively), touching circle ABCD} Therefore, since FG touches circle ABCD, and EA has been joined from the center E to the point of contact A, the angles at A are thus right-angles [Prop. 3.18]. So, for the same (reasons), the angles at points B, C, and D are also right-angles. And since angle AEB is a right- angle, and EBG is also a right-angle, GH is thus parallel to AC [Prop. 1.29]. So, for the same (reasons), AC is also parallel to FK. So that GH is also parallel to FK [Prop. 1.30]. So, similarly, we can show that GF and HK are each parallel to BED. Thus, GK, GC, AK, FB, and BK are (all) parallelograms. Thus, GF is equal to HK, and GH to FK [Prop. 1.34]. And since AC is equal to BD, but AC (is) also (equal) to each of GH and FK, and BD is equal to each of GF and HK [Prop. 1.34] [and each of GH and FK is thus equal to each of GF and HK], the quadrilateral FGHK is thus equilateral. So I say that (it is) also right-angled. For since GBEA is a parallelogram, and AEB is a right-angle, AGB is thus also a right-angle [Prop. 1.34]. So, similarly, we can show that the angles at H, K, and F are also right-angles. Thus, FGHK is right-angled. And it was also shown (to be) equilateral. Thus, it is a square [Def. 1.22]. And it has been circumscribed about circle ABCD. Thus, a square has been circumscribed about the given circle. (Which is) the very thing it was required to do. t See the footnote to the previous proposition. * See the footnote to Prop. 3.34. 116 ETOIXEIfiN 5'. ELEMENTS BOOK 4 Etc to 6oi9£v TeTpdywvov xuxXov syYpd^ai. "Eaxco to BotJev xsxpdycovov to ABrA. 8a 5r| sic; to ABrA TETpdywvov xuxXov eyYpd^ai. A E A K B TeT[if]o , d« exaTepa tcov AA, AB 8i)(a xaTa Ta E, Z arista, xal Sid (lev tou E ouoTspa tcov AB, TA TiapdXXrjXoc; r^/iSio 6 E9, 8id 8s tou Z orcoTspa xwv AA, Br TtapdXXrjXoc; fj/iSw f) ZK' uapaXXr)X6Ypa|ji[j.ov dpa eaTiv exacrcov xSv AK, KB, A9, 6A, AH, Hr, BH, HA, xod ai aTtervavTiov ai)TG)v TiXeupal StjXovoti I'aai [eiaiv]. xal snel lar\ sgtIv f) AA Tfj AB, xa[ ecrci xfj<; [lev AA r^iiaeia f] AE, Tfjc; 6e AB f\[iiaeiaL f\ AZ, lar\ dpa xal f\ AE Tfj AZ' wots xal ai am- vavTiov i'ar) dpa xal f\ ZH Tfj HE. 6^io[m<; 8r] Bd^o^ierv, oti xal exaTepa t£>v H0, HK exaTepa twv ZH, HE eaTiv Xary ai Teaaapsc dpa ai HE, HZ, H0, HK laai dXXiqXaii; [eiaiv]. 6 dpa xevTpo \ism tw H SiaaT^aTi 8e evl xuv E, Z, 0, K xuxXoc; YP°tcp6^ievo<; fj^ei xal 8id tGv Xomwv ar^eiwv xal ecpdijiSTai twv AB, Br, TA, A A euiDeiwv 8id to opiDdt; eivai Tag npbc, Toiz E, Z, <d, K Y^viac;' ei yap tejisI 6 xuxXoc; Tag AB, Br, TA, AA, f] Tfj 8ia|jieTpcp tou xuxXou rcpoc; opi&dc; caz dxpac; dyoiievr) evto<; TieaELTai tou xuxXou' onep aTOTtov sSei/i}/). oux dpa 6 xevTptp tG H BiaaTr^aTi 8s fevi iwv E, Z, 0, K xuxXog ypacpojisvog tsjiei Tag AB, Br, TA, AA su-deiag. ecpd(|ieTai dpa auT&v xal eaTai eYYSYpa\i\ievoci si<z to ABrA TETpaYWvov. Ei<; dpa to Bo-dev TeTpdyMvov xuxXog EYY£YP a7txal ' onep s8ei Ttoirjaai. ft'. nspi to Bo'dev T£TpdY«vov xuxXov nepiYpd^ai. TiaTCO to Boftsv TCTpaYwvov to ABrA- BsT Br) xcepl to ABrA TSTpdYWvov xuxXov nxpiYpd^ai. Proposition 8 To inscribe a circle in a given square. Let the given square be ABCD. So it is required to inscribe a circle in square ABCD. A E D K B H C Let AD and AB each have been cut in half at points E and F (respectively) [Prop. 1.10]. And let EH have been drawn through E, parallel to either of AB or CD, and let FK have been drawn through F, parallel to either of AD or BC [Prop. 1.31]. Thus, AK, KB, AH, HD, AG, GC, BG, and GD are each parallelograms, and their opposite sides [are] manifestly equal [Prop. 1.34]. And since AD is equal to AB, and AE is half of AD, and AF half of AB, AE (is) thus also equal to AF. So that the opposite (sides are) also (equal). Thus, FG (is) also equal to GE. So, similarly, we can also show that each of GH and GK is equal to each of FG and GE. Thus, the four (straight- lines) GE, GF, GH, and GK [are] equal to one another. Thus, the circle drawn with center G, and radius one of E, F, H, or K, will also go through the remaining points. And it will touch the straight-lines AB, BC, CD, and DA, on account of the angles at E, F, H, and K being right-angles. For if the circle cuts AB, BC, CD, or DA, then a (straight-line) drawn at right-angles to a diameter of the circle, from its extremity, will fall inside the circle. The very thing was shown (to be) absurd [Prop. 3.16]. Thus, the circle drawn with center G, and radius one of E, F, H, or K, does not cut the straight-lines AB, BC, CD, or DA. Thus, it will touch them, and will have been inscribed in the square ABCD. Thus, a circle has been inscribed in the given square. (Which is) the very thing it was required to do. Proposition 9 To circumscribe a circle about a given square. Let ABCD be the given square. So it is required to circumscribe a circle about square ABCD. 117 ETOIXEIfiN 5'. ELEMENTS BOOK 4 'ETU^eux'detaai yap ai Ar, BA xe^ivexcoaav dXXr|Xa<; xaxa to E. A Kai end Tar) eaxlv f| AA xfj AB, xoivf] Se f) Ar, 860 Sf] ai AA, Ar Sua! xdi<; BA, Ar I'aai eiaiv xal pdau; f) Ar pdaei xfj Br Xor\- ywvia dpa f] uko AAr yovia xfj utio BAr Tar] laxiv f] dpa 0tc6 AAB ytovia Bi^a xex^trjxai utio xfjc; Ar. 6[io[«<; 8f) 8eie;ouev, 6x1 xal exdaxr] xc5v Gtco ABr, BrA, TAA 5ixa xex^rjxai utio xfiv Ar, AB eMeiCSv. xal CTtel iar] eaxlv f) utio AAB ywvia xfj utio ABr, xa( eaxi xfj? uev utio AAB f)uiaeia f) utio EAB, xfjc; Se 0tc6 ABr f]\j.iaeia r] utio EBA, xal f) Gtco EAB dpa xfj utio EBA eaxiv ten)' £>axe xal TiXeupa f) EA xfj EB eaxiv lay). b[io'ux>c, Sf] 8eic;o[iev, 6x1 xal exaxepa xwv EA, EB [euT!)eiGv] exaxepa x£Sv Er, EA lay] eaxiv. ai xeaaapec; dpa ai EA, EB, Er, EA Taai dXXrjXaic; eiaiv. 6 dpa xevxpw ifi E xal Siaaxrj^taxi evl iwv A, B, r, A xuxXoc; ypacpojievoc; fjc;ei xal Sid xwv Xoitiwv arj^ieiwv xal eaxai Tiepiyeypa[i[ievoc; Ttepl xo ABrA xexpdywvov. Tiepiyeypdcp'dw «<; 6 ABrA. nepl xo 8oi9ev dpa xexpdywvov xuxXoc TiepiyeypaTixar OTtep e8ei Tioirjaai. l . 'IaoaxeXec; xpiywvov auaxf|aaaiL>ai e^ov exaxepav xwv Tcpog xfj pdaei ywvifiv 8iTiXaa[ova xfjc; Xomfj^ . 'Exxeia'dw xu; euiJeTa f) AB, xal xex[if]adw xaxa xo r a/][ieIov, waxe xo utio xwv AB, Br Tiepiexouevov opiDoywviov laov eTvai x£> arco xfjc; TA xexpaycovw' xal xevxpo xc5 A xal 8iaaxf|[iaxi xw AB xuxXoc; yeypdcpif)w 6 BAE, xal evr)p[i6adw eic; xov BAE xuxXov xfj Ar eMeia [ly) [xeiCovi ouarj xfjc; xou BAE xuxXou 8ia[iexpou lot] eu-dela f] BA- xal CTieCeux'dwaav ai AA, Ar, xal Tiepiyeypdcp'dw Ttepl xo ArA xpiywvov xuxXoc; 6 ArA. AC and BD being joined, let them cut one another at E. A E c And since DA is equal to AB, and AC (is) common, the two (straight-lines) DA, AC are thus equal to the two (straight-lines) BA, AC. And the base DC (is) equal to the base BC. Thus, angle DAC is equal to angle BAC [Prop. 1.8]. Thus, the angle DAB has been cut in half by AC. So, similarly, we can show that ABC, BCD, and CD A have each been cut in half by the straight-lines AC and DB. And since angle DAB is equal to ABC, and EAB is half of DAB, and EBA half of ABC, EAB is thus also equal to EBA. So that side EA is also equal to EB [Prop. 1.6]. So, similarly, we can show that each of the [straight-lines] EA and EB are also equal to each of EC and ED. Thus, the four (straight-lines) EA, EB, EC, and ED are equal to one another. Thus, the circle drawn with center E, and radius one of A, B, C, or D, will also go through the remaining points, and will have been circumscribed about the square ABCD. Let it have been (so) circumscribed, like ABCD (in the figure). Thus, a circle has been circumscribed about the given square. (Which is) the very thing it was required to do. Proposition 10 To construct an isosceles triangle having each of the angles at the base double the remaining (angle) . Let some straight-line AB be taken, and let it have been cut at point C so that the rectangle contained by AB and BC is equal to the square on CA [Prop. 2.11]. And let the circle BDE have been drawn with center A, and radius AB. And let the straight-line BD, equal to the straight-line AC, being not greater than the diame- ter of circle BDE, have been inserted into circle BDE [Prop. 4.1]. And let AD and DC have been joined. And let the circle ACD have been circumscribed about trian- gle ACD [Prop. 4.5]. 118 ETOIXEIfiN 5'. ELEMENTS BOOK 4 Kal etceI to utco x65v AB, Br Taov eaxi x£> duo xfjc AT, iar] 8s f] Ar xrj BA, to dpa utio xuv AB, Br taov eaxl xc5 dn:6 xfj<; BA. xod ercel xuxXou xou ArA ei'Xr]Tcxa[ xi ar^ieTov exxoc; xo B, xal dmo xou B Ttp6<; xov ArA xuxXov TcpoaTccTcxtoxaai 8uo eu-delai ai BA, BA, xal f] [lev auxcov xe^vei, f) Be: TtpoaTimxei, xa[ eaxi xo utco xwv AB, Br Taov xw (xko xfjc BA, f] BA apa ecpdacxexai xou ArA xuxXou. CTcei ouv ecp&Tcxexai [iev f] BA, duo 8e xrjc xaxd xo A enacprjg 8irjxxai f) Ar, f] apa Otto BAr ycovid for) eoxl xrj ev xfi evaXXdc; xou xuxXou xpir|[iaxi yiovia xrj utio AAr. eTiei ouv Iar] eaxiv f) utio BAr xrj utco AAr, xoivr] Tcpoaxeia'fko rj Otto TAA- 6Xr) apa f) utio BAA Tar] eaxl 8ual xau; utio TAA, AAr. dXXa xalg Otio TAA, AAr lot] eaxiv f] exxo<; f) utio BTA- xal f] Otto BAA apa iar) eaxl xrj utio BrA. dXXa rj Otto BAA xrj Otto TBA eaxiv lor), end xal TiXeupa rj AA xrj AB eaxiv 'iar}- uaxs xal f] Otio ABA xrj Otio BrA eaxiv far), ai xpdc apa ai Otio BAA, ABA, BrA laai dXXr|Xai<; eiaiv. xal excel i'ar) saxlv f) utio ABr yovia xrj utio BrA, Xar] eaxl xal TtXeupd f] BA TiXeupa xrj Ar. dXXa r) BA xrj TA Oiroxeixai iar]- xal f] TA apa xrj TA eaxiv Xaf]- waxe xal yovia r] Otto TAA yovia xrj utio AAr eaxiv iar]' ai apa utco TAA, AAr xfj<; utio AAr eiai 8iTtXaaiou<;. Xar] Se f] utco BrA xau; utco TAA, AAr- xal f] utio BrA apa xfj<; utio TAA eaxi BiTcXfj. iar) 8e r] utio BrA exaxepa xwv utio BAA, ABA- xal exaxepa apa xwv Otio BAA, ABA xfjc utco AAB eaxi BiTcXfj. 'IaoaxeXet; apa xpiywvov auveaxaxai xo ABA e/ov exaxepav xwv Tcpoc; xrj AB pdaei ywvifiv BiTcXaaiova xfj<; Xomrjc OTcep e8ei Tioirjaai. ia'. EE? xov 8o$evxa xuxXov Tievxdywvov iaoTcXeupov xe xal And since the (rectangle contained) by AB and BC is equal to the (square) on AC, and AC (is) equal to BD, the (rectangle contained) by AB and BC is thus equal to the (square) on BD. And since some point B has been taken outside of circle ACD, and two straight- lines BA and BD have radiated from B towards the cir- cle ACD, and (one) of them cuts (the circle), and (the other) meets (the circle), and the (rectangle contained) by AB and BC is equal to the (square) on BD, BD thus touches circle ACD [Prop. 3.37]. Therefore, since BD touches (the circle), and DC has been drawn across (the circle) from the point of contact D, the angle BDC is thus equal to the angle DAC in the alternate segment of the circle [Prop. 3.32]. Therefore, since BDC is equal to DAC, let CD A have been added to both. Thus, the whole of BDA is equal to the two (angles) CD A and DAC. But, the external (angle) BCD is equal to CD A and DAC [Prop. 1.32]. Thus, BDA is also equal to BCD. But, BDA is equal to CBD, since the side AD is also equal to AB [Prop. 1.5]. So that DBA is also equal to BCD. Thus, the three (angles) BDA, DBA, and BCD are equal to one another. And since angle DBC is equal to BCD, side BD is also equal to side DC [Prop. 1.6]. But, BD was assumed (to be) equal to CA. Thus, CA is also equal to CD. So that angle CD A is also equal to angle DAC [Prop. 1.5]. Thus, CD A and DAC is double DAC. But BCD (is) equal to CD A and DAC. Thus, BCD is also double CAD. And BCD (is) equal to to each of BDA and DBA. Thus, BDA and DBA are each double DAB. Thus, the isosceles triangle ABD has been con- structed having each of the angles at the base BD double the remaining (angle). (Which is) the very thing it was required to do. Proposition 11 To inscribe an equilateral and equiangular pentagon 119 ETOIXEIfiN 5'. ELEMENTS BOOK 4 iaoycoviov eyypdtjiai. "Eaxw 6 Bo'dek; xuxXoc; 6 ABrAE- 8eT 8r) ei<; xov ABrAE xuxXov Tievxdycovov iaoTtXeupov xe xal iaoycjviov eyypdtjiai.. 'Exxdcrdw xpiytovov iaoaxeXec to ZH0 8inXaaiova e)(ov exaxepav xGv Tipoc; xou; H, ywviaiv xrjc Tipog xcp Z, xai eyyeypdcpiEko ei<; xov ABrAE xuxXov xfi ZH6 xpiywvw iaoyciiviov xpiycovov to AEA, &axe xfj [lev ixpog xto Z yojvia Tarjv eTvai xrjv utio TAA, exaxepav 6e xc5v Tipoc; xolc; H, i'or)v exaxepa x«v utio ArA, TAA- xal exaxepa dpa xfiv utco ArA, TAA xfjc utco TAA eaxi SiTiXrj. xexjnqcdt) 8r| exaxepa xfiv utio ArA, TAA Stya Otco exaxepac; x«v TE, AB eMeioSv, xal eiceCeux^woav ai AB, Br, AE, EA. 'Eiiei ouv exaxepa xcov utio ArA, TAA ycoviwv 6l- TcXaaicov eaxl xfjc; uico TAA, xal xexjrrjjievai eial 8i)(a utio xfiv TE, AB euiSeicov, ai itevxe dpa ycoviai ai utio AAr, ArE, EEA, TAB, BAA I'oai dXXf)Xai<; eiaiv. ai 8e laai ywviai era latov nepicpepeuov pepiqxaaLv ai nevxe dpa nz- picpepeiai ai AB, Br, EA, AE, EA laai dXXrjXaic; eiaiv. utio 8e xdc laac; Tcepicpepeiac; i'oai euifteTai UTcoxeivouaiv ai itevxe dpa euOelai ai AB, Br, TA, AE, EA laai dXXf|Xaic; eiaiv iaoTtXeupov dpa eoxi xo ABrAE Tievxdycovov. Xeyto 8iq, oxi xal iaoytoviov. enel yap f) AB Tcepicpepeia xfj AE tcs- pi9epeia eaxlv Tar), xoivr) Tcpoaxeiai^co f) BrA- oXr) dpa f] ABrA nepicpepia 6Xr) xfj EArB nepicpepeia eaxlv i'or). xal PePrjxev era ^ev xfjg ABrA nepicpepeiac ycovia f] utco AEA, era Se xfjc EArB Tcepicpepeiac; ywvia f] utio BAE- xal f) utio BAE dpa yiovia xrj utio AEA eaxiv la/]. 8id xd auxd 8f) xal exdaxr) x£5v utio ABr, BrA, TAE ycovicov exaxepa xSv utco BAE, AEA eaxiv lory iaoycoviov dpa eaxl xo ABrAE Tievxdywvov. eSeiy^r) Se xal iaonXeupov. Eic; dpa xov Bcdevxa xuxXov nevxayiovov iaonXeupov xe xal iaoycbviov eyyeypanxai- onep e8ei noifjaai. IP'. nepi xov 8oi9evxa xuxXov Tievxdycovov iaoTcXeupov xe xai iaoycoviov Ttepiypdtjiai. in a given circle. Let ABCDE be the given circle. So it is required to inscribed an equilateral and equiangular pentagon in cir- cle ABCDE. Let the the isosceles triangle FGH be set up hav- ing each of the angles at G and H double the (angle) at F [Prop. 4.10]. And let triangle ACD, equiangular to FGH, have been inscribed in circle ABCDE, such that CAD is equal to the angle at F, and the (angles) at G and H (are) equal to ACD and CD A, respectively [Prop. 4.2]. Thus, ACD and CD A are each double CAD. So let ACD and CD A have been cut in half by the straight-lines CE and DB, respectively [Prop. 1.9]. And let AB, BC, DE and EA have been joined. Therefore, since angles ACD and CD A are each dou- ble CAD, and are cut in half by the straight-lines CE and DB, the five angles DAC, ACE, ECD, CDB, and BDA are thus equal to one another. And equal angles stand upon equal circumferences [Prop. 3.26]. Thus, the five circumferences AB, BC, CD, DE, and EA are equal to one another [Prop. 3.29]. Thus, the pentagon ABCDE is equilateral. So I say that (it is) also equiangular. For since the circumference AB is equal to the circumfer- ence DE, let BCD have been added to both. Thus, the whole circumference ABCD is equal to the whole cir- cumference EDCB. And the angle AED stands upon circumference ABCD, and angle BAE upon circumfer- ence EDCB. Thus, angle BAE is also equal to AED [Prop. 3.27]. So, for the same (reasons), each of the an- gles ABC, BCD, and CDE is also equal to each of BAE and AED. Thus, pentagon ABCDE is equiangular. And it was also shown (to be) equilateral. Thus, an equilateral and equiangular pentagon has been inscribed in the given circle. (Which is) the very thing it was required to do. Proposition 12 To circumscribe an equilateral and equiangular pen- tagon about a given circle. 120 ETOIXEIfiN 5'. ELEMENTS BOOK 4 K r A "Eax« 6 Bodeic xuxXoc 6 ABrAE- 8eT Be Tiepi xov ABrAE xuxXov Tievxdyovov EaoTiXeupov xe xal iaoyoviov Tiepiypdijiai. Nevorfo'dti) xoO eyyeypa^evou Tievxaywvou xwv ytoviwv ar)^eTa xd A, B, T, A, E, coaxe lane, elvai xdc AB, Br, TA, AE, EA Tiepicpepeiac- xal Sid xwv A, B, V, A, E f]X"Q(^oay xoO xuxXou ecpoaixo^ievoa ai H0, 9K, KA, AM, MH, xal eiXr]tp , dw xou ABrAE xuxXou xevxpov xo Z, xal STieCeuxtiuaav ai ZB, ZK, ZT, ZA, ZA. Kal ETiel f] ^iev KA eu$eTa ecpdrcxexai xoO ABrAE xaxd xo T, arco Be xoO Z xevxpou era xf)v xaxd xo T eTiacpf)v CTieCeuxxai f) Zr, f] Zr dpa xdiJexoc eaxiv Ira xf)v KA- op-df) dpa eaxiv exaxepa xwv Tipoc iu T ytovicov. Bloc xd auxd 8f) xal ai Tipoc xolc B, A ar)fieioic ytoviai op-dai eiaiv. xal end opiDr] eaxiv f) utio ZrK y wvia, xo dpa arco xfjc ZK foov eaxi xolc aKO xov Zr, TK. 8id xd auxd 8f) xal xolc duo xwv ZB, BK i'aov laxl xo duo xfjc ZK- waxe xd aTio xwv Zr, TK xolc duo tuv ZB, BK eaxiv foa, Sv xo duo xfjc Zr iw dfto xfjc ZB eaxiv i'aov Xoitiov dpa xo diio xfjc TK xo diio xfjc BK eaxiv i'aov. for] dpa f] BK xfj TK. xal euel for] eaxlv f] ZB xrj Zr, xal xoivf] r] ZK, 8uo Bf] ai BZ, ZK Bual xdic rZ, ZK foai eiaiv xai pdaic f] BK pdaei xrj TK [eaxiv] fory ycovia dpa f] ^tev utio BZK [ycovia] xrj Otio KZT eaxiv for)- f] Be utio BKZ xrj utio ZKr- SmXfj dpa f] ^tev utio BZr xfjc utio KZr, f) 8e utio BKr xfjc utio ZKr. Bid xd auxd 8f] xai f] ^iev utio TZA xfjc utio TZA eaxi BuiXfj, fj 8e utio AAr xfjc utio ZAr. xal STid for] eaxlv f) Br Tiepicpepeia xrj TA, tar] eaxi xal ywvia f] utio BZr xrj utio TZA. xai eaxiv f) [Lev utio BZr xfjc utio KZr 8iTiXfj, f] Be utio AZr xfjc Otio AZr- for] dpa xal f) utio KZr xfj utio AZE eaxi 8e xal f] utio ZrK ytovia xfj utio ZrA i'ar). Suo Sf] xpiyovd eaxi xd ZKr, ZAr xdc Suo ywviac xdic Suai ywviaic foac exovxa xal h'iolv TiXeupdv uia TiXeupa iar]v xoivf]v auxGv xf]v ZE xal xdc XoiTidc dpa TiXeupdc xdic XoiTidic TiXeupdic i'aac ecei xal xf)v XoiTif]v ywviav xfj Xomfj ywvia- i'ar) dpa f] [lev KT eO'dsTa xfj TA, f) 8e utio ZKr ywvia xfj utio ZAr. xai etisI iar) eaxlv f) Kr xfj TA, SmXfj dpa f) KA xfjc Kr. Bid xd auxa Bf) 8ei)(Tf)f]a£xai xal f) 8K xfjc BK BiTiXfj. xai eaxiv f) BK xfj Kr iar)- xal f) 9K dpa xfj KA eaxiv for). 6(ioitoc 8f) Beix'driaexai G K C L Let ABCDE be the given circle. So it is required to circumscribe an equilateral and equiangular pentagon about circle ABCDE. Let A, B, C, D, and E have been conceived as the an- gular points of a pentagon having been inscribed (in cir- cle ABCDE) [Prop. 3.11], such that the circumferences AB, BC, CD, DE, and EA are equal. And let GH, BK, KL, LM, and MG have been drawn through (points) A, B, C, D, and E (respectively), touching the circled And let the center F of the circle ABCDE have been found [Prop. 3.1]. And let FB, FK, FC, FL, and FD have been joined. And since the straight-line KL touches (circle) ABCDE at C, and FC has been joined from the center F to the point of contact C, FC is thus perpendicular to KL [Prop. 3.18]. Thus, each of the angles at C is a right- angle. So, for the same (reasons), the angles at B and D are also right-angles. And since angle FCK is a right- angle, the (square) on FK is thus equal to the (sum of the squares) on FC and CK [Prop. 1.47]. So, for the same (reasons), the (square) on FK is also equal to the (sum of the squares) on FB and BK. So that the (sum of the squares) on FC and CK is equal to the (sum of the squares) on FB and BK, of which the (square) on FC is equal to the (square) on FB. Thus, the remain- ing (square) on CK is equal to the remaining (square) on BK. Thus, BK (is) equal to CK. And since FB is equal to FC, and FK (is) common, the two (straight- lines) BF, FK are equal to the two (straight-lines) CF , FK. And the base BK [is] equal to the base CK. Thus, angle BFK is equal to [angle] KFC [Prop. 1.8]. And BKF (is equal) to FKC [Prop. 1.8]. Thus, BFC (is) double KFC, and BK C (is double) FKC. So, for the same (reasons), CFD is also double CFL, and DLC (is also double) FLC. And since circumference BC is equal to CD, angle BFC is also equal to CFD [Prop. 3.27]. And BFC is double KFC, and DFC (is double) LFC. Thus, KFC is also equal to LFC. And angle FCK is also equal to FCL. So, FKC and FLC are two triangles hav- 121 ETOIXEIfiN 5'. ELEMENTS BOOK 4 xal exdaxr] xcov 0H, HM, MA exaxepa xcov 0K, KA icny laoTiXeupov apa eaxl to H0KAM Ttevxdycovov. Xeyco 5t|, oxi xal laoycoviov. enel yap for) eaxlv f\ Gtco ZKr ycovia xfj utio ZAr, xal eSeix^T) xfjc \ism Gtco ZKT BmXrj f\ utio 0KA, xfj? 8s utio ZAr 8mXrj f] utio KAM, xal f] 0tc6 0KA apa xfj utio KAM eaxiv Tar), o^oicoc 8f) Beiyjdriaexai xal exdaxr) xcov utio K6H, OHM, HMA exaxepa xcov uno 0KA, KAM larj- ai Tievxe apa ycovlai ai utio H0K, 6KA, KAM, AMH, MH0 laai aXXriXaic eiaiv. laoycoviov apa eaxl xo H9KAM Tievxdycovov. e8s[)(i9r) Se xal laoTcXEupov, xal Tcepiyeypanxai Tiepl xov ABrAE xuxXov. [Ilepl xov Scdevxa apa xuxXov TCEvxaycovov loonXeupov xe xal laoycoviov TtspiysypaTixai] • onep eBa Tioifjaai. ing two angles equal to two angles, and one side equal to one side, (namely) their common (side) FC. Thus, they will also have the remaining sides equal to the (cor- responding) remaining sides, and the remaining angle to the remaining angle [Prop. 1.26]. Thus, the straight-line KC (is) equal to CL, and the angle FKC to FLC. And since KC is equal to CL, KL (is) thus double KC. So, for the same (reasons), it can be shown that H K (is) also double BK. And BK is equal to KC. Thus, HK is also equal to KL. So, similarly, each of HG, CM, and ML can also be shown (to be) equal to each of HK and KL. Thus, pentagon GHKLM is equilateral. So I say that (it is) also equiangular. For since angle FKC is equal to FLC, and HKL was shown (to be) double FKC, and KLM double FLC, HKL is thus also equal to KLM. So, similarly, each of KHG, HGM, and GML can also be shown (to be) equal to each of HKL and KLM. Thus, the five angles GHK, HKL, KLM, LMG, and MGH are equal to one another. Thus, the pentagon GHKLM is equiangular. And it was also shown (to be) equilateral, and has been circumscribed about circle ABCDE. [Thus, an equilateral and equiangular pentagon has been circumscribed about the given circle]. (Which is) the very thing it was required to do. t See the footnote to Prop. 3.34. If. Fic, xo 8oil>£v Tcevxdyovov, 6 saxiv taoTiXeupov xe xal laoycoviov, xuxXov Eyypd^ai. A r k a "Eaxco xo BoiJev Ttevxdycovov laoTtXeupov xe xal laoycovi- ov xo ABrAE- 6el 6f] tic, xo ABrAE Tcevxdycovov xuxXov eyypdtjiai. Texuf]a , dco yap exaxepa xcov utio BIA, TAE ycovicov 8[)(a utco exaxepa? xcov TZ, AZ eu'deicov xal aTco xou Z arjudou, xaO' o aujipdXXouaiv dXXrjXau; ai TZ, AZ euiMai, eTte^eu)cdcoaav ai ZB, ZA, ZE eui9elai. xal eTcel Tar) eaxlv Proposition 13 To inscribe a circle in a given pentagon, which is equi- lateral and equiangular. A C K D Let ABCDE be the given equilateral and equiangular pentagon. So it is required to inscribe a circle in pentagon ABCDE. For let angles BCD and CDE have each been cut in half by each of the straight-lines CF and DF (re- spectively) [Prop. 1.9]. And from the point F, at which the straight-lines CF and DF meet one another, let the 122 ETOIXEIfiN 5'. ELEMENTS BOOK 4 f) Br xfj TA, xoivf] Be f] TZ, 860 8f) ai Br, TZ Suol xaic; Ar, rZ iaai eiaiv xal yiovia f) (mo BrZ ycovia tfi bub ArZ [eaxiv] for) - pdaic; dpa f] BZ pdaei xfj AZ eaxiv iar], xal xo BrZ xpiywvov xw ArZ xpiywvw eaxiv laov, xal ai Xoinai ycoviai xaic; XomaTc; ycoviaic; laai eaovxai, ucp' ac al laai rcXeupal ujraxeivouaiv for) apa f] (mo TBZ ycovia xfj O716 TAZ. xal en:el BiuXfj eaxiv f] Otco IAE xfjc; bub IAZ, Xat] 8e f] (jlsv Otto TAE xfj Otto ABr, f] Se uko TAZ xfj hub TBZ, xal f] uuo TBA dpa xfjc; utco TBZ eaxi SiitXfj- for] apa f] utco ABZ ycovia xfj \mb ZBE f] apa Otto ABr ycovia 8[)(a xex^rjxai Otto xfjc; BZ sbftemc,. o^toicoc; Bf] 8ei)(iL>r)aexai, oxi xal exaxepa xcov bub BAE, AEA Bi)(a xex^rjxai bub exaxepac; xcov ZA, ZE e0i!)eic5v. fj)(Tf)waav 8f] duo xou Z ar^eiou etxI xac; AB, Br, TA, AE, EA eOiJeiac; xd-dexoi ai ZH, Z0, ZK, ZA, ZM. xal creel tar] eaxiv f) Otio 9rZ ycovia xfj bnb KFZ, eaxi 8e xal opiDf] f) Oreo Z8r [6pi!)fj] xfj Oreo ZKr for), 860 8f] xpiycovd eaxi xd Z6r, ZKr xac; 860 ycoviac; 8ual ycoviaic; laac; exovxa xa ' t^ av TtXeupdv TcXeupa iar)v xoivfjv auxcov xf)v Zr Orcoxeivouaav bnb [iiav xcov lacov ycovicov xal xac; XoiTtdc; apa TtXeupac; xaic; XoiTcdic; TiXeupdic; laac; '€E,zv Tar) apa f) Z9 xdiitexoc; xf) ZK xai9exco. ojioicoc; 8f) SeivOfjaexai, oxi xal exdaxr) xcov ZA, ZM, ZH exaxepa xcov Z0, ZK for] eaxiv ai rcevxe dpa eO-delai ai ZH, Z6, ZK, ZA, ZM foai dXXfjXaic; eiaiv. 6 apa xevxpco xco Z 8iaaxf][iaxi 8e evi xcov H, 0, K, A, M xuxXoc; ypacpojievoc; fjc;ei xal 81a xwv Xoitccov arj^ieicov xal ecpdij^exai xwv AB, Br, TA, AE, EA e0i5eicov 81a xo opiSdc; eTvai xac; rcpoc; xou; H, 0, K, A, M ar^eioic; ycoviac;. si yap oOx e(pd<}>exai aOxcov, dXXd xe^iel auxdc;, aujj.pf]aexai xf]v xfj 8ia[iexpco xou xuxXou npoc; op-ddc; die' axpac; dyojievrjv evxoc; TUTtxeiv xou xuxXou' oTiep axoKov eBeivdr). oux apa 6 xsvxpw xfi Z 8iaaxf]^.axi 8e evi xcov H, &, K, A, M arjpieiwv ypacpo^tevoc; xuxXoc; xe^tel xac; AB, Br, EA, AE, EA eb-deiac,- ecpdi^exai apa auxfiv. yeypdcpi^w wc; 6 H0KAM. Eic; apa xo So-dev Tievxdyovov, 6 eaxiv iaoiiXeupov xe xal iaoyoviov, xuxXoc; eyyeypanxai- onep e8ei uoifjaai. i5'. nepl xo 8oiL>ev iievxdywvov, o eaxiv iaonXeupov xe xal iaoycoviov, xuxXov ixepiypd^ai. ^axo xo 8oT5ev uevxdywvov, 6 eaxiv iaoTiXeupov xe xal straight-lines FB, FA, and FE have been joined. And since BC is equal to CD, and CF (is) common, the two (straight-lines) BC, CF are equal to the two (straight- lines) DC, CF. And angle BCF [is] equal to angle DCF. Thus, the base BF is equal to the base DF, and triangle BCF is equal to triangle DCF, and the remain- ing angles will be equal to the (corresponding) remain- ing angles which the equal sides subtend [Prop. 1.4]. Thus, angle CBF (is) equal to CDF. And since CDF is double CDF, and CDE (is) equal to ABC, and CDF to CBF, CBA is thus also double CBF. Thus, angle ABF is equal to F BC. Thus, angle ABC has been cut in half by the straight-line BF. So, similarly, it can be shown that BAE and AED have been cut in half by the straight-lines FA and FE, respectively. So let FG, FH, FK, FL, and FM have been drawn from point F, per- pendicular to the straight-lines AB, BC, CD, DE, and EA (respectively) [Prop. 1.12]. And since angle HCF is equal to KCF, and the right-angle FHC is also equal to the [right-angle] FKC, FHC and FKC are two tri- angles having two angles equal to two angles, and one side equal to one side, (namely) their common (side) FC, subtending one of the equal angles. Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26]. Thus, the perpendicular FH (is) equal to the perpendicular FK. So, similarly, it can be shown that FL, FM, and FG are each equal to each of FH and FK. Thus, the five straight-lines FG, FH, FK, FL, and FM are equal to one another. Thus, the circle drawn with center F, and radius one of G, H, K, L, or M, will also go through the remaining points, and will touch the straight-lines AB, BC, CD, DE, and EA, on account of the angles at points G, H, K, L, and M being right-angles. For if it does not touch them, but cuts them, it follows that a (straight-line) drawn at right- angles to the diameter of the circle, from its extremity, falls inside the circle. The very thing was shown (to be) absurd [Prop. 3.16]. Thus, the circle drawn with center F, and radius one of G, H, K, L, or M, does not cut the straight-lines AB, BC, CD, DE, or EA. Thus, it will touch them. Let it have been drawn, like GHKLM (in the figure). Thus, a circle has been inscribed in the given pen- tagon which is equilateral and equiangular. (Which is) the very thing it was required to do. Proposition 14 To circumscribe a circle about a given pentagon which is equilateral and equiangular. Let ABCDE be the given pentagon which is equilat- 123 ETOIXEIftN 5'. ELEMENTS BOOK 4 iaoycoviov, to ABEAE- 8sT 8f) itepi to ABrAE TtsvTdycovov xuxXov Tiepiypd(J>ai. A TeT[if]a , dco §r) sxaTspa twv Otto BrA, EAE yioviSSv Sixa Otto sxaT£pa<; t£>v TZ, AZ, xod arco tou Z ar)(jisiou, xa-d' o aujipdXXouoiv ai Eui&Elai, Em. Ta B, A, E ar]\isia £Tie^£U)fd«aav sMsTai ai ZB, ZA, ZE. ojioiwc; 8/) tG Ttpo toutou 6ei)cdr]0£Tai, oti xod sxaaTT) twv Otto TBA, BAE, AEA yiovifiv 8[)(a T£T|ir]Tai Otto exdoTT)<; tGv ZB, ZA, ZE eut}eiGv. xod ettel Tor) eotIv r) Otto BTA ycovia Tfj Otto EAE, xai egti Tfj<; ^iev Otto BrA f][iia£ia r) Otto ZEA, Tfjg 8s Otto EAE f^iasia f] Otto EAZ, xod r] Otto ZrA dpa Tfj Otto ZAr egtiv icy]- Sots xod nXsupd f) ZT TiXsupa Tfj ZA eotlv lor). by.o'\.oic, 8r) 8ei)cdr]0£Tai, oti xod exdoTT) tGv ZB, ZA, ZE sxaTspa t«v Zr, ZA saw tar)- ai tievts dpa su-daai ai ZA, ZB, Zr, ZA, ZE i'oai dXXfjXaig siaiv. 6 dpa xsvTpcp tw Z xai 8iaaTr]jj.aTi svi t£Sv ZA, ZB, ZT, ZA, ZE xuxXoc; ypacpojisvoc; ffesi. xal 8id t«v Xoitmov a/jjisiwv xod eaxoti tee- piysypajijjisvoc;. TtEpiysypdcp^co x °d eo"tm 6 ABrAE. ILepi dpa to 8ot5ev TiEVTdyojvov, 6 eotiv iaoTtXsupov te xal iaoycoviov, xuxXoc TCEpiysypaTtTai' ousp eSel Tioirjaai. is'. El? tov 8oiD£VTa xuxXov s^dyiovov iaonXsupov te xal iaoycoviov Eyypdtjiai. 'EaTCO 6 So-dslc; xuxXo<; 6 ABEAEZ- SeT Sf] sic; tov ABrAEZ xuxXov sc;dycovov iaoTtXsupov te xai iaoycoviov Eyypdtjjai. "H)(Tf)w toO ABrAEZ xuxXou 8id^iSTpo<; f) AA, xai siXficp'dco to XEVTpov toO xuxXou to H, xai xsvTpcp jisv tw A 8iaaTT]^aTi 8s iu AH xuxXoc; ysypdcpiJco 6 EHr0, xai £7u£EU)cdsTaai ai EH, TH Sifix^coaav ski Ta B, Z ar)[i£la, xai ETCEc'Eu/'dcoaav ai AB, Br, TA, AE, EZ, ZA- Xsyco, oti eral and equiangular. So it is required to circumscribe a circle about the pentagon ABCDE. A So let angles BCD and CDE have been cut in half by the (straight-lines) CF and DF, respectively [Prop. 1.9]. And let the straight-lines FB, FA, and FE have been joined from point F, at which the straight-lines meet, to the points B, A, and E (respectively). So, similarly, to the (proposition) before this (one), it can be shown that angles CBA, BAE, and AED have also been cut in half by the straight-lines FB, FA, and FE, respec- tively. And since angle BCD is equal to CDE, and FCD is half of BCD, and CDF half of CDE, FCD is thus also equal to FDC. So that side FC is also equal to side FD [Prop. 1.6]. So, similarly, it can be shown that FB, FA, and FE are also each equal to each of FC and FD. Thus, the five straight-lines FA, FB, FC, FD, and FE are equal to one another. Thus, the circle drawn with center F, and radius one of FA, FB, FC, FD, or FE, will also go through the remaining points, and will have been circumscribed. Let it have been (so) circumscribed, and let it be ABCDE. Thus, a circle has been circumscribed about the given pentagon, which is equilateral and equiangular. (Which is) the very thing it was required to do. Proposition 15 To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle. So it is required to inscribe an equilateral and equiangular hexagon in circle ABCDEF. Let the diameter AD of circle ABCDEF have been drawn, t and let the center G of the circle have been found [Prop. 3.1]. And let the circle EGCH have been drawn, with center D, and radius DC And EG and CG being joined, let them have been drawn across (the cir- 124 ETOIXEIfiN 5'. ELEMENTS BOOK 4 to ABrAEZ s^aytovov iaonXeupov xe eaxi xal iaoyioviov. e A Tkei yap xo H ar^elov xevxpov sail xou ABrAEZ xuxXou, igt] eaxlv f) HE xrj HA. raiXiv, end to A a/jjiaov xevxpov sail toO Hr0 xuxXou, for) eaxlv r) AE xrj AH. dXX' f\ HE xrj HA eSelx'dr) tar) - xal f] HE apa xrj EA tar) eaxlv iaoTtXeupov apa eaxi xo EHA xplycovov xal ai xpeu; apa auxou yoviai ai bnb EHA, HAE, AEH laai dXXr]Xai<; eiaiv, C7iei8r]TT;ep xwv EaoaxsXwv xpiycivov ai Kpo<; xrj pdaei ywviai laai dXXrjXaic eialv xal eiaiv ai xpeu; xou xpiywvou ywviai Buaiv op^aTc; low f] apa urab EHA ywvla xplxov eaxl 860 6pif)Gv. 6^iolo<; 5r] Seix'driaexai xal f\ bnb AHT xplxov 860 op^wv. xal ckcI f] TH eu'deTa em xf]v EB axa'delaa xa<; ecpe£rj<; ywvla<; xa<; bnb EHr, THB Sualv opiJau; laac noieT, xal Xomrj apa f] O716 THB xplxov eaxl 860 op-dGv ai apa bnb EHA, AHr, THB ywvlai laai dXXr]Xai<; eialv oaxe xal ai xaxa xopucprjv auxdic ai bnb BHA, AHZ, ZHE i'aai eialv [xdlc bnb EHA, AHr, THB]. ai e? apa ywvlai ai (mo EHA, AHr, THB, BHA, AHZ, ZHE I'aai dXX^Xai.; eiaiv. ai 8e laai ywvlai era lawv Ttepicpepeiwv peprjxaaiv al ec; apa Tte- picpepeiai ai AB, Br, EA, AE, EZ, ZA i'aai dXXr]Xaic eialv. bnb 8e xa<; laac; Ttepicpepelac al laai eu-deTai UTtoxeivouaiv ai §5 apa eu'delai laai dXXf|Xai<; eiaiv iaorcXeupov apa eaxl xo ABrAEZ eQaywvov. Xeyw §7), 6x1 xal iaoywviov. emci yap lot] eaxlv f] ZA Ttepicpepeia xrj EA Ttepicpepeia, xoivr] TipoaxelaiDw f] ABrA Ttepicpepeia- oXr] apa f) ZABrA oXr] xrj EArBA eaxiv I'ar)- xal peprjxev era ^tev xfj? ZABrA Ttepicpepeia<; f] bnb ZEA ywvla, era 8e xfj? EArBA Ttepi- cpepelac; rj bnb AZE yovia - iar] apa f) tmo AZE ywvia xfj bnb AEZ. o^iolcoc; 8r) 8ei)edr|aexai, 6x1 xal ai XoiTtai ywvlai xou ABrAEZ e^aywvou xaxa piiav laai eiaiv exaxepa xwv bnb AZE, ZEA ywviwv iaoywviov apa eaxl xo ABrAEZ e^dywvov. eSeix^r] 8e xal iaoTtXeupov xal eyyeypauxai eic; xov ABrAEZ xuxXov. Eiz apa xov BoiJevxa xuxXov e^dywvov iaoTtXeupov xe cle) to points B and F (respectively). And let AB, BC, CD, BE, EF, and FA have been joined. I say that the hexagon ABCDEF is equilateral and equiangular. H A For since point G is the center of circle ABCDEF, GE is equal to CD. Again, since point D is the cen- ter of circle GCH, DE is equal to DC. But, GE was shown (to be) equal to GD. Thus, GE is also equal to ED. Thus, triangle EGD is equilateral. Thus, its three angles EGD, GDE, and DEG are also equal to one an- other, inasmuch as the angles at the base of isosceles tri- angles are equal to one another [Prop. 1.5]. And the three angles of the triangle are equal to two right-angles [Prop. 1.32]. Thus, angle EGD is one third of two right- angles. So, similarly, DGC can also be shown (to be) one third of two right-angles. And since the straight-line CG, standing on EB, makes adjacent angles EGG and CGB equal to two right-angles [Prop. 1.13], the remain- ing angle CGB is thus also one third of two right-angles. Thus, angles EGD, DGC, and CGB are equal to one an- other. And hence the (angles) opposite to them BGA, AGF, and FGE are also equal [to EGD, DGC, and CGB (respectively)] [Prop. 1.15]. Thus, the six angles EGD, DGC, CGB, BGA, AGF, and FGE are equal to one another. And equal angles stand on equal cir- cumferences [Prop. 3.26]. Thus, the six circumferences AB, BC, CD, DE, EF, and FA are equal to one an- other. And equal circumferences are subtended by equal straight-lines [Prop. 3.29]. Thus, the six straight-lines {AB, BC, CD, DE, EF, and FA) are equal to one another. Thus, hexagon ABCDEF is equilateral. So, I say that (it is) also equiangular. For since circumfer- ence FA is equal to circumference ED, let circumference ABCD have been added to both. Thus, the whole of FABCD is equal to the whole of EDCBA. And angle FED stands on circumference FABCD, and angle AFE on circumference EDCBA. Thus, angle AFE is equal 125 ETOIXEIfiN 5'. ELEMENTS BOOK 4 xal iaoywviov EYY^YP a7txal ' OTI£ P e8ei noifjaai. I16pia[Jia. 'Ex 5r) xouxou (pavepov, oxi f) xoO e^aywvou TtXeupa tar) saxl xrj ex xoO xevxpou xoO xuxXou. Ojioicdc; 8e xou; era tou Tievxaycbvou eav 8ia xcov xaxa xov xuxXov Siaipeaewv ecpaTtxojievac; xou xuxXou aYayt^ev, TxspiYpacpr^aexai Ttepl xov xuxXov e^aYWvov laoTtXeupov xe xal iooywviov dxoXoui&Mt; xou; ski xoO TievxaY^vou eipr^evoig. xal exi Bia xwv o^ioicov xou; etu xou nevxaYCovou eipr^evoi? sic, xo Bcdsv s^otYCOvov xuxXov eYyp6n\>o[Lev t£ xal TiepiYpa^o^tEv oTiep e5ei Ttoirjaai. t See the footnote to Prop. 4.6. El? xov ScfJevxa xuxXov TisvxexaiBexaYCOvov laonXsupov xe xal EaoYCoviov tyYpcn\)ai. A Tiaxo 6 Bo-deu; xuxXo? 6 ABIA- SeT Br] sic, xov ABrA xuxXov nevxExaiSexaYCovov laoitXeupov xe xal Iooycoviov 'EyysyP^V'S" £ k TOV ABrA xuxXov xpiY«vou \iev tao- TiXeupou xoO zlc, auxov eYYP°«po^evou TtXeupa f) Ar, Ttev- xaYWvou 8s laoiiXeupou f\ AB - oiwv apa eaxlv 6 ABrA xuxXoc Tacov xp^axwv Bexaravxe, xoiouxmv f] [ie\ ABT Txepicpepeia xptxov ouaa xou xuxXou eaxai Ttevxe, f) Be AB Txepicpepeia Tte^txov ouaa xou xuxXou eaxai xpiwv Xomf] apa to DEF [Prop. 3.27]. Similarly, it can also be shown that the remaining angles of hexagon ABC DEF are in- dividually equal to each of the angles AFE and FED. Thus, hexagon ABC DEF is equiangular. And it was also shown (to be) equilateral. And it has been inscribed in circle ABCDE. Thus, an equilateral and equiangular hexagon has been inscribed in the given circle. (Which is) the very thing it was required to do. Corollary So, from this, (it is) manifest that a side of the hexagon is equal to the radius of the circle. And similarly to a pentagon, if we draw tangents to the circle through the (sixfold) divisions of the (cir- cumference of the) circle, an equilateral and equiangu- lar hexagon can be circumscribed about the circle, analo- gously to the aforementioned pentagon. And, further, by (means) similar to the aforementioned pentagon, we can inscribe and circumscribe a circle in (and about) a given hexagon. (Which is) the very thing it was required to do. Proposition 16 To inscribe an equilateral and equiangular fifteen- sided figure in a given circle. A Let ABCD be the given circle. So it is required to in- scribe an equilateral and equiangular fifteen-sided figure in circle ABCD. Let the side AC of an equilateral triangle inscribed in (the circle) [Prop. 4.2], and (the side) AB of an (in- scribed) equilateral pentagon [Prop. 4.11], have been in- scribed in circle ABCD. Thus, just as the circle ABCD is (made up) of fifteen equal pieces, the circumference ABC, being a third of the circle, will be (made up) of five 126 ETOIXEIfiN 5'. ELEMENTS BOOK 4 f] Br xov Tacov 860. xexjnqcrdGj f\ Br Sixa xaxa to E- sxaxspa apa xfiiv BE, Er TCpicpspeifiv ravxexaiSexaxov eaxi xou ABrA xuxXou. 'Eav apa £Tu£euc;avxec; Tag BE, Er I'aac; auxau; xaxa to auvsy^ su-deiac; svap^oawjisv sic; xov ABrA[E] xuxXov, eaxai tic, auxov £yy£YP°W£ vov TievxExaiBexdycovov iaorcXeu- pov xe xal laoyoviov ojtep eSei Ttoirjaai.. 'Ojioiwc; 8e xou; era xou nevxaYWvou eav 81a xfiv xaxa xov xuxXov 8iaipeaewv ecpauxo^svac; xou xuxXou dyaYW^ev, TtepiYpacprjaexai Ttepl xov xuxXov Ttevxexai- 8exaYWvov laoitXeupov xe xal iooywviov. erxi 8e Sid xwv ojioiwv xolc; era xou nsvxaYWvou Seli;e«v xal etc; xo Bo'dev TievxexaiBexaYWvov xuxXov eyy p6n\to[ie\i T £ xal ne- piYpdtjjo^iev oiiep eSei Ttoirjaai.. such (pieces), and the circumference AB, being a fifth of the circle, will be (made up) of three. Thus, the remain- der BC (will be made up) of two equal (pieces). Let (cir- cumference) BC have been cut in half at E [Prop. 3.30]. Thus, each of the circumferences BE and EC is one fif- teenth of the circle ABODE. Thus, if, joining BE and EC, we continuously in- sert straight-lines equal to them into circle ABCD[E] [Prop. 4.1], then an equilateral and equiangular fifteen- sided figure will have been inserted into (the circle). (Which is) the very thing it was required to do. And similarly to the pentagon, if we draw tangents to the circle through the (fifteenfold) divisions of the (cir- cumference of the) circle, we can circumscribe an equilat- eral and equiangular fifteen-sided figure about the circle. And, further, through similar proofs to the pentagon, we can also inscribe and circumscribe a circle in (and about) a given fifteen-sided figure. (Which is) the very thing it was required to do. 127 128 ELEMENTS BOOK 5 Proportion tThe theory of proportion set out in this book is generally attributed to Eudoxus of Cnidus. The novel feature of this theory is its ability to deal with irrational magnitudes, which had hitherto been a major stumbling block for Greek mathematicians. Throughout the footnotes in this book, a, ft, 7, etc., denote general (possibly irrational) magnitudes, whereas m, n, I, etc., denote positive integers. 129 ETOIXEIfiN z. ELEMENTS BOOK 5 "Opoi. a'. Mepog taxi (ieyeiL>oc; \Leye r &o\jQ xo IXaaoov xou [idZovoz, oxav xaxajiexpfj to \is%o\. P'. IToXXaTtXaoiov 8s to jieT^ov xou eXdxxovoc, oxav xa- xatiexprjxai Otto xou eXdxxovoc;. y'. Aoyog eoxl 860 y.sys-&&\i ojioyevSiv f] xaxd tltj- Xixox/jxd n:oia a/eaic;. 8'. Aoyov l)(£iv Tipoc; aXXr)Xa (ieye'dr) Xeyexai, a Suvaxai TtoXXaTiXaoia^ojjeva dXXf|Xiov UTtepe/eiv. e'. 'Ev xw auxC) Xoyw jieye'dr] Xeyexai etvai TtpGxov Tipoc; 8euxepov xal xpixov Tipoc; xexapxov, oxav xd xou Tipcixou xa[ xpixou lodxic; TioXXaTiXdoia x«v xou 8euxepou xal xexdpxou iodxic; KoXXanXaaicov xaif)' otioiovouv TioXXa- TiXaoiaojiov exdxepov exaxepou fj ajja bnzpixT] °V a l ' CTa fl fj djjia eXXeiTifj Xrjcpdevxa xaxdXX/jXa. 9'. Td Se xov auxov e/ovxa Xoyov (jieyeiSr) dvdXoyov xaXeurdco. C. "Oxav Se xfiv iodxic; TioXXaTiXaoitov xo jjcv xou Tipcoxou TtoXXanXdoiov UTiepexn TO ° TO ° Beuxepou tioX- XaTiXaoiou, xo 8e xou xpixou TioXXaTiXdaiov [ir\ UTiepexT) xou xou xexdpxou TioXXaTiXaaiou, xoxe xo npwxov npbc, xo Beuxepov jjeii^ova Xoyov e^eiv Xeyexai, fjTiep xo xpixov Tipoc; xo xexapxov. /)'. AvaXoyia 8e ev xpiolv opoig ekayioTf] eoxiv. •&'. "Oxav Be xpia jieye'dr] dvdXoyov fj, xo Tipoxov Tipoc xo xpixov BmXaoiova Xoyov e/eiv Xeyexai f)7tep Tipoc; xo Seuxepov. 1'. "Oxav 8e xeooapa (icye^T) dvdXoyov fj, xo Ttpcoxov Tipoc; xo xexapxov xpmXaoiova Xoyov e)(eiv Xeyexai fjTiep Tipoc; xo Beuxepov, xal del tEjff, ojjlolcoc;, 6? av f] dvaXoyia UTidpxr). ia'. 'OjioXoya (leyeiSr] Xeyexai xd jjcv f]youjieva xou; f]you^evoi<; xd 8e CTtojjeva xolc; enojievoic. iP'. 'EvaXXd^ Xoyog eoxl Xfjcjxc; xou fjyoujjevou Tipoc; xo fjyoujievov xal xou ercojjievou Tipoc; xo eixojievov. iy'. AvaTiaXiv Xoyog eoxl Xfjcjiu; xou enojievou tbc; fjyou^evou Tipoc; xo fjyoufjievov cbc; ctiojjcvov. 18'. Euvdeoic; Xoyou eoxl Xrjcjnc; xou fjyou^ievou (iexd xou enojievou (be evoc Tipoc; auxo xo euojievov. ie'. Aiaipeou; Xoyou eoxl Xfjcjiic xfjc UTiepoxfjc;, fj unepe/ei xo f)you(ievov xou enojievou, Tipoc; auxo xo e7i6[ievov. 19'. 'Avaoxpocpr) Xoyou eoxl Xfj(J>ic; xou f)you(ievou npoc xrjv U7iepo);r]v, f) unepe/ei xo fjyoujievov xou C7io|jievou. iC- Ai° ioou Xoyoc; eoxl nXeiovwv ovxtov [ieycdGv xal dXXwv auxolg Tocov xo nX^oc ouvSuo Xajj.pavojj.evwv xal ev x£S auxw Xoyco, oxav fj &>c, ev xolc; upcoxoig jjeye-deoi xo npSxov Tipoc; xo eo^axov, ouxwc; ev xolg Seuxepoic; jjeye-deoi xo npSxov Tipog xo eo/axov fj aXXwg- Xfj^it; x£5v axpcov Definitions 1. A magnitude is a part of a(nother) magnitude, the lesser of the greater, when it measures the greater.^ 2. And the greater (magnitude is) a multiple of the lesser when it is measured by the lesser. 3. A ratio is a certain type of condition with respect to size of two magnitudes of the same kind.-f 4. (Those) magnitudes are said to have a ratio with re- spect to one another which, being multiplied, are capable of exceeding one another. § 5. Magnitudes are said to be in the same ratio, the first to the second, and the third to the fourth, when equal multiples of the first and the third either both exceed, are both equal to, or are both less than, equal multiples of the second and the fourth, respectively, being taken in corre- sponding order, according to any kind of multiplication whatever. 1 6. And let magnitudes having the same ratio be called proportional.* 7. And when for equal multiples (as in Def. 5), the multiple of the first (magnitude) exceeds the multiple of the second, and the multiple of the third (magnitude) does not exceed the multiple of the fourth, then the first (magnitude) is said to have a greater ratio to the second than the third (magnitude has) to the fourth. 8. And a proportion in three terms is the smallest (possible). 8 9. And when three magnitudes are proportional, the first is said to have to the third the squaredll ra tio of that (it has) to the second.^ 10. And when four magnitudes are (continuously) proportional, the first is said to have to the fourth the cubedW ratio of that (it has) to the second. §§ And so on, similarly, in successive order, whatever the (continuous) proportion might be. 11. These magnitudes are said to be corresponding (magnitudes): the leading to the leading (of two ratios), and the following to the following. 12. An alternate ratio is a taking of the (ratio of the) leading (magnitude) to the leading (of two equal ratios), and (setting it equal to) the (ratio of the) following (mag- nitude) to the following. ^ 13. An inverse ratio is a taking of the (ratio of the) fol- lowing (magnitude) as the leading and the leading (mag- nitude) as the following.** 14. A composition of a ratio is a taking of the (ratio of the) leading plus the following (magnitudes), as one, to the following (magnitude) by itself. $$ 130 ETOIXEIfiN z. ELEMENTS BOOK 5 xocd' UKS^aipeoiv xwv (irotov. 15. A separation of a ratio is a taking of the (ratio it)'. Tsxapaypievr] 8e dvaXoyia eoxiv, oxav xpiGv ovxwv of the) excess by which the leading (magnitude) exceeds ^eyei^Gv xod dXXwv auxolc; icxov xo nXfj'doc yivrjxai <i>c; [iev the following to the following (magnitude) by itself. I sv xoic; Tipcoxoic; [Leye-Qeoiv f]youjj.£vov Ttpoc; sttojisvov, ouxgjc 16. A conversion of a ratio is a taking of the (ratio sv xoic; Beuxepou; ^Leyen&Eaiv f]you^i£vov Tipoc; etto^evov, tbg of the) leading (magnitude) to the excess by which the 8e ev ioXc, upcoxoic; jisys-decnv etto^evov npbc, dXXo xi, ouxwg leading (magnitude) exceeds the following. W sv xolc; Seuxepou; dXXo xi Tipoc rjyou^evov. 17. There being several magnitudes, and other (mag- nitudes) of equal number to them, (which are) also in the same ratio taken two by two, a ratio via equality (or ex aequali) occurs when as the first is to the last in the first (set of) magnitudes, so the first (is) to the last in the sec- ond (set of) magnitudes. Or alternately, (it is) a taking of the (ratio of the) outer (magnitudes) by the removal of the inner (magnitudes). 18. There being three magnitudes, and other (magni- tudes) of equal number to them, a perturbed proportion occurs when as the leading is to the following in the first (set of) magnitudes, so the leading (is) to the following in the second (set of) magnitudes, and as the following (is) to some other (i.e., the remaining magnitude) in the first (set of) magnitudes, so some other (is) to the leading in the second (set of) magnitudes. §§§ t In other words, a is said to be a part of (3 if f3 = m a. t In modern notation, the ratio of two magnitudes, a and (3, is denoted a : (3. § In other words, a has a ratio with respect to f3 if m a > f3 and n/3 > a, for some m and n. ' In other words, a : /3 : : 7 : <5 if and only if m a > n/3 whenever m 7 > n S, and ma = n/3 whenever m 7 = n 8, and ma < n/3 whenever rn 7 < n 8, for all m and n. This definition is the kernel of Eudoxus' theory of proportion, and is valid even if a, (3, etc., are irrational. * Thus if a and (3 have the same ratio as 7 and <5 then they are proportional. In modern notation, a : (3 :: 7 : <5. $ In modern notation, a proportion in three terms — a, f3, and 7 — is written: a : f3 :: f3 : 7. II Literally, "double". tt In other words, if a : (3 :: (3 : 7 then a : 7 :: a 2 : (3 2 . H Literally, "triple". In other words, if a : (3 :: f3 : 7 7 : <5 then a : 8 :: a 3 : f3 3 . ■ ■ In other words, if a : (3 :: 7 : <5 then the alternate ratio corresponds to a : 7 :: (3 : 8. ** In other words, if a : (3 then the inverse ratio corresponds to f3 : a. $$ In other words, if a : (3 then the composed ratio corresponds to a + (3 : f3. II II In other words, if a : (3 then the separated ratio corresponds to a — f3 : f3. ttt In other words, if a : (3 then the converted ratio corresponds to a : a — (3. it* In other words, if a, (3, 7 are the first set of magnitudes, and 5, e, ( the second set, and a : f3 : 7 :: 8 : e : then the ratio via equality (or ex aequali) corresponds to a : 7 :: <5 : ( . §§§ In other words, if a, (3, 7 are the first set of magnitudes, and <5, e, ( the second set, and a : (3 :: 8 : e as well as (3 : 7 :: ( : 8, then the proportion is said to be perturbed. a'. Proposition V 'Eav rj onoaaouv jisys-d/) otcogcovouv jieye^wv Tacov xo If there are any number of magnitudes whatsoever TzkyjOoQ excxgxov ex&oxou iadxi<; ttoXXootMchov, oacmXdaiov (which are) equal multiples, respectively, of some (other) saxiv ev x«v ^sys'dcSv smoq, xooauxaTiXdoia eaxou xal xd magnitudes, of equal number (to them), then as many 131 ETOIXEIfiN z. ELEMENTS BOOK 5 A H B T A i 1 1 i 1 1 Ei 1 Z' 1 TSaxw OTOaaouv ^ey^dr] xd AB, FA otcogcovoOv [ls- ys-dcov twv E, Z Tacov to TtXrjiSoc; ixaaxov exaaxou iadxig TtoXXanXdaiov Xeyco, oxi oaaitXdaiov laxi xo AB xou E, xoaauxanXdaia scroti xal xd AB, TA xwv E, Z. EtceI ydp Eadxic ecm KoXXa^Xdoiov xo AB xou E xal xo TA xou Z, oaa dpa eaxlv ev xG AB ^.eyea}/] iaa x£> E, xoaauxa xal ev xw TA Taa xo Z. 8ir)pf|crd« xo [lev AB eic xd to E (ieye'dr) laa xd AH, HB, xo 8e TA eic. xd t£> Z laa xd TO, 0A- saxai 8r) laov xo TiXfj-dog xfiv AH, HB t£> TtXr|iL>£i tGv T6, 9A. xal STieUaov eotI to [ism AH t« E, to 8e T6 tw Z, '(gov dpa to AH t« E, xal Ta AH, TO toTc E, Z. Sloc Ta auTa Sr] laov sotI to HB tw E, xal Ta HB, 0A tou; E, Z- oaa dpa eotIv ev to AB I'oa t£> E, ToaauTa xal ev toTc AB, TA I'aa toTc E, Z- ooaTiXdoiov dpa lav. to AB tou E, TooauTauXdoia saTai xal Ta AB, TA tcov E, Z. 'Edv dpa f) oTiooaouv ^sysdr) otiooovouv [leyei&Gv I'owv to nkfjdoz exaoTov sxdoTou iadxic noXXanXdaiov, oaaitXdaiov eaTiv ev tcov jisys'daiv evoc, TooauTanXdoia eaTai xal Ta ndvTa twv itdvTiov oitsp e8ei SeT^ai. t In modern notation, this proposition reads m a + m f3 H P'- 'Edv upcoTov SeuTepou iadxic f) uoXXanXdoiov xal Tprcov TeTapTou, f] 8e xal kc^ititov SeuTepou iadxic TtoXXaitXdaiov xal Ixtov TCTapTou, xal auvTSiDev TtpCrcov xal TtejiTtTov SeuTepou iadxic saTai noXXaTiXdoiov xal Tprcov xal sxtov TSTdpTOU. np«Tov ydp to AB BeuTEpou tou T iadxic soto tcoX- XanXdoiov xal Tphov to AE TSTapTou toO Z, egtw 8e xal ne[LKTOv to BH SeuTepou tou T iadxic TtoXXaTtXdaiov xal sxtov to EG TSTapTou tou Z - Xeyw, oti xal auvTEiDev npfiTov xal TCEjiTtTov to AH SeuTepou tou T iadxic eaxoa TioXXanXdaiov xal TpiTov xal extov to A9 TETapTou tou Z. times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second) . A G B C H D i 1 1 i 1 1 Ei 1 F' 1 Let there be any number of magnitudes whatsoever, AB, CD, (which are) equal multiples, respectively, of some (other) magnitudes, E, F, of equal number (to them) . I say that as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F. For since AB, CD are equal multiples of E, F, thus as many magnitudes as (there) are in AB equal to E, so many (are there) also in CD equal to F. Let AB have been divided into magnitudes AG, GB, equal to E, and CD into (magnitudes) CH, HD, equal to F. So, the number of (divisions) AG, GB will be equal to the num- ber of (divisions) CH, HD. And since AG is equal to E, and CH to F, AG (is) thus equal to E, and AG, CH to E, F. So, for the same (reasons), GB is equal to E, and GB, HD to E, F. Thus, as many (magnitudes) as (there) are in AB equal to E, so many (are there) also in AB, CD equal to E, F. Thus, as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F. Thus, if there are any number of magnitudes what- soever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisi- ble) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second) . (Which is) the very thing it was required to show. Proposition X If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (mag- nitude) and a sixth (are) also equal multiples of the sec- ond and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and the sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively) . For let a first (magnitude) AB and a third DE be equal multiples of a second C and a fourth F (respec- tively). And let a fifth (magnitude) BG and a sixth EH also be (other) equal multiples of the second C and the fourth F (respectively) . I say that the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, = m{a + 13 + ■■■). 132 ETOIXEIfiN z. ELEMENTS BOOK 5 A B H i 1 1 1 1 1 r i — i A E i 1 1 1 1 1 Z' 1 'Etc! yap iodxu; eax! TtoXXaitXdoiov to AB xou T xa! to AE toO Z, ooa dpa eoxlv ev xG AB Toa xG T, xooauxa xal ev xG AE 10a xG Z. Sia xd auxd Sr) xal ooa eoxlv ev iS BH I'aa xG T, xooauxa xal ev xG EG 10a xG Z- ooa dpa eoxlv ev 6X9 xG AH 10a xG T, xooauxa xal ev oXtp xG A9 10a xG Z- ooarcXdoiov dpa eoxl xo AH xou T, xooauxarcXdoiov eoxai xal xo AQ xou Z. xal ouvxei9ev dpa TtpGxov xal tc^itixov xo AH Seuxepou xou T iodxu; eoxai TtoXXaitXdoiov xal xpixov xal exxov xo A6 xexdpxou xou Z. 'Edv dpa TipGxov Seuxepou lodxic; fj TioXXarcXaoiov xal xpixov xexdpxou, fj Se xal nejiTixov Seuxepou iodxu; ttoX- XanXdoiov xal exxov xexdpxou, xal ouvxcdev TtpGxov xal tc^tixov Seuxepou lodxic; eoxai TioXXaTtXdoiov xal xpixov xal exxov xexdpxou- oitep eSei Sel^ai. t In modern notation, this propostion reads m a + n a = (m + n) a. Y- 'Edv TipGxov Seuxepou lodxic; fj TioXXaiiXdoiov xal xpixov xexdpxou, X/]cpi9fj Se lodxic; TioXXaiiXdoia xou xe upGxou xal xpixou, xal Si' loou xGv Xrjqj'devxwv exdxepov exaxepou lodxic; eoxai TioXXajiXdaiov xo [Lev xou Seuxepou xo Se xou xexdpxou. npGxov Y a P TO A Seuxepou xou B lodxic; eoxw tioX- XajiXdaiov xal xpixov xo T xexdpxou xou A, xal eiXr]cp , dw xGv A, T lodxic; uoXXaTiXdaia xa EZ, H0- Xeyw, oxi lodxic; eoxl TioXXauXdoiov xo EZ xou B xal xo H9 xou A. 'Etc! yap lodxic; eoxl KoXXanXdaiov xo EZ xou A xal xo H0 xou r, ooa dpa eoxlv ev xG EZ loa xG A, xooauxa xal ev xG H9 I'oa xG T. Sir)pf]o , dw xo ^ev EZ zlc, xd xG A ^leyei!)/] I'oa xd EK, KZ, xo Se H0 zlc, xd xG T i'oa xa HA, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively) . A B G i 1 1 1 1 1 C' — ' D EH i 1 1 1 1 1 F i 1 For since AB and DE are equal multiples of C and F (respectively), thus as many (magnitudes) as (there) are in AB equal to C, so many (are there) also in DE equal to F. And so, for the same (reasons), as many (magnitudes) as (there) are in BG equal to C, so many (are there) also in EH equal to F. Thus, as many (magnitudes) as (there) are in the whole of AG equal to C, so many (are there) also in the whole of DH equal to F. Thus, as many times as AG is (divisible) by C, so many times will DH also be divisible by F. Thus, the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively) . Thus, if a first (magnitude) and a third are equal mul- tiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magni- tude) and the fifth, being added together, and the third and sixth, (being added together), will also be equal mul- tiples of the second (magnitude) and the fourth (respec- tively) . (Which is) the very thing it was required to show. Proposition 3 f If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multi- ples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively. For let a first (magnitude) A and a third C be equal multiples of a second B and a fourth D (respectively), and let the equal multiples EF and GH have been taken of A and C (respectively). I say that EF and GH are equal multiples of B and D (respectively). For since EF and GH are equal multiples of A and C (respectively), thus as many (magnitudes) as (there) are in EF equal to A, so many (are there) also in GH 133 ETOIXEIfiN z. ELEMENTS BOOK 5 AO- eaxai 8r] I'aov to Tikf(Qoc, tcov EK, KZ t£3 7iXr|iL>£i xfiv HA, AO. xal inel lodxic; eaxl TioXXaTrXdaiov to A tou B xal to T tou A, laov 8e to [ie\ EK t« A, to 8s HA tG T, iadxic; dpa taxi TroXXaTtXaaiov to EK tou B xal to HA tou A. Sid Ta auTa 8/) iadxi<; sotl TroXXaTrXdaiov to KZ tou B xal to A9 tou A. end ouv npfiTov to EK 8euTepou tou B I'odxic; eotI TroXXaTtXaaiov xal TpiTov to HA TSTapTou tou A, ecrci 8e xal nejiTCTov to KZ SeuTepou tou B lodxig TroX- XaTtXaaiov xal extov to A6 TSTapTou tou A, xal auvTE^ev dpa upcdTov xal TtefiTrcov to EZ SsuTspou tou B iadxic; eotI TroXXaTtXaaiov xal TpiTov xal sxtov to HO TETapTou tou A. A' 1 1 1 B 1 E K Z i 1 1 ri — i — i — i A' — ' HA© i 1 1 Edv dpa TtpCrtov BeuTEpou laaxic; fj TroXXaTtXaaiov xal TpiTov TSTapTou, X/]cpi9fj 8e tou TtpcVtou xal TpiTou iadxic; TtoXXaTtXdaia, xal 8i° I'aou tGv X/jcpi&svTiov sxaTspov sxaTspou iadxic; eaTai TroXXaTtXaaiov to tou 8euTepou to 8s tou TSTapTou- oTtep SSei 8eic;ai. t In modern notation, this proposition reads m(n a) = (m n) a. 5'. 'Edv 7ip£>T0v Ttpoc; 6euTepov tov auTov zyr\ Xoyov xal Tpaov Ttpoc; TETapTov, xal Ta iadxu; TtoXXaitXdaia tou te upcoTou xal TpiTou Ttpoc; Ta iadxic; TtoXXaTtXdaia tou 8euTepou xal TSTapTou xai}° ottoiovouv TtoXXaTtXaaiaajjiov tov auTov e^ei Xoyov Xrjcp^evTa xaTaXXrjXa. npfiTov yap to A Ttpoc; SeuTepov to B tov auTov eyixto Xoyov xal xpixov to T Ttpoc; xexapxov to A, xal eiXrjtp'dw xfiv [ie\ A, T iadxic; TtoXXaTtXdaia Ta E, Z, t«v 8e B, A dXXa, a exuxev, iadxic; TtoXXaTtXdaia Ta H, 6- Xsyw, oxi saxlv cbc; to E Ttpoc; to H, ouxcoc; to Z Ttpoc; to 6. equal to C. Let EF have been divided into magnitudes EK, KF equal to A, and GH into (magnitudes) GL, LH equal to C. So, the number of (magnitudes) EK, KF will be equal to the number of (magnitudes) GL, LH. And since A and C are equal multiples of B and D (re- spectively), and EK (is) equal to A, and GL to C, EK and GL are thus equal multiples of B and D (respec- tively). So, for the same (reasons), KF and LH are equal multiples of B and D (respectively) . Therefore, since the first (magnitude) EK and the third GL are equal mul- tiples of the second B and the fourth D (respectively), and the fifth (magnitude) KF and the sixth LH are also equal multiples of the second B and the fourth D (re- spectively), then the first (magnitude) and fifth, being added together, (to give) EF, and the third (magnitude) and sixth, (being added together, to give) GH, are thus also equal multiples of the second (magnitude) B and the fourth!) (respectively) [Prop. 5.2]. A I 1 1 1 B' 1 E K F i 1 1 C i — " — i — i D' 1 G L H i 1 1 Thus, if a first (magnitude) and a third are equal mul- tiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal mul- tiples of the second (magnitude) and the fourth, respec- tively. (Which is) the very thing it was required to show. Proposition 4 f If a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, be- ing taken in corresponding order, according to any kind of multiplication whatsoever. For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D. And let equal multiples E and F have been taken of A and C (respectively), and other random equal multiples G and 134 ETOIXEIfiN z. ELEMENTS BOOK 5 A' 1 B i-H Ei 1 1 H 1 — i — i — 1 K' 1 ' Mi 1 1 1 r i — i Zi — i — i A, , , N' ' ' ' EiXrjcp'dco yap xfiiv (lev E, Z iadxu; TtoXXanXdaia xd K, A, t«v 5e H, dXXa, a exuxev, iadxig TioXXanXdoia xd M, N. [Kod] fejtel tadxig eaxl TtoXXanXdaiov to jisrv E xou A, xo 8e Z xou r, xal eiXr)7ixai xfiv E, Z I'odxic; noXXanXdoia xd K, A, ladxic dpa sraxl noXXanXdoiov xo K xou A xal xo A xou T. Bid xd auxd 8r| ladxi? eaxl KoXXanXdaiov xo M xou B xal xo N xou A. xal inei eaxiv cbc xo A upoc; xo B, oux«<; xo T 7ip6<; xo A, xal EiXrjTtxai xov ^tev A, T ladxic; TtoXXaTtXdaia xd K, A, xQv 8e B, A dXXa, a exu)(£v, iadxu; TtoXXaTtXdaia xd M, N, ei dpa UTtepe)(£i xo K xou M, UTtepexei xocl xo A xou N, xal si laov, laov, xal et eXaxxov, eXaxxov. xal saxi xd [Ltv K, A xQv E, Z iadxi<; noXXauXdaia, xd 8e M, N xwv H, dXXa, a exu)(£v, iadxic; noXXaKXaaia' saxiv dpa foe, xo E Ttpoc xo H, oux«<; xo Z Tipoc xo 0. 'Edv dpa upwxov Kpbc, Ssuxepov xov auxov i)(T] Xoyov xal xpixov 7tp6<; xexapxov, xal xd ladxu; noXXanXdaia xou xe Tipoxou xal xptxou npbz xd ladxu; TioXXauXdaia xou Seuxepou xal xexdpxou xov auxov e^ei Xoyov xai}' okoiovouv uoXXa- TiXaaiao^tov XTjcpiJevxa xaxdXXrjXa- ojisp eSsi 8el5ai. H of B and D (respectively) . I say that as E (is) to G, so F (is) to H. A ' ' B i—i E i 1 1 G i — i — i — i K i 1 1 Mi 1 1 1 C i — i D ' ' F i 1 1 H L i 1 1 N i 1 1 1 For let equal multiples K and L have been taken of E and F (respectively), and other random equal multiples M and N of G and H (respectively) . [And] since E and F are equal multiples of A and C (respectively), and the equal multiples K and L have been taken of E and F (respectively), K and L are thus equal multiples of A and C (respectively) [Prop. 5.3]. So, for the same (reasons), M and N are equal multiples of B and D (respectively). And since as A is to B, so C (is) to D, and the equal multiples K and L have been taken of A and C (respectively), and the other random equal multiples M and N of B and D (respectively), then if K exceeds M then L also exceeds N, and if (K is) equal (to M then L is also) equal (to N), and if {K is) less (than M then L is also) less (than N) [Def. 5.5]. And K and L are equal multiples of E and F (respectively), and M and N other random equal multiples of G and H (respectively) . Thus, as E (is) to G, so F (is) to H [Def. 5.5]. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multi- ples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : (3 : : 7 : $ then m a : n /3 : : m 7 : n <5, for all m and n. 135 ETOIXEIfiN z. ELEMENTS BOOK 5 s . 'Eav [liyE'doz (leye'doug iadxic; fj TtoXXaTtXdaiov, onep dcpaips-dev dcpaips-devToc;, xal to Xolttov tou Xomou iadxic; soxai TroXXaTtXdcHov, oacmXdmov soti to oXov tou oXou. A E B i 1 1 1 — i — i — i H r Z A I 1 1 1 Meye'doc yap to AB [leye-Qouc, tou IA iadxi<; sgtco TtoX- XoaiXdaiov, orcep d(paipe , d£v to AE dcpaipe'dEVToc; tou TZ' Xeyw, oti xal Xoitiov to EB Xoitiou tou ZA Eadxic; eaxoa TtoXXaitXaaiov, oaomXdaiov eaTiv oXov to AB oXou tou IA. 'OaanXdaiov ydp taxi to AE tou TZ, ToaauTaaiXdaiov ysyovsTW xal to EB tou TH. Kal snel ladxn; ecttI TtoXXaTtXdaiov to AE tou TZ xal to EB tou Hr, iadxi<; apa sotI TtoXXaTtXdaiov to AE tou TZ xal to AB tou HZ. xeroxi Be iadxii; uoXXaTiXdoiov to AE tou TZ xal to AB tou IA. ladxig apa sgti noXXauXdaiov to AB ExaTepou twv HZ, EA igov apa to HZ iu IA. xoivov a(pr}p7)a-Q(x> to TZ- Xoiitov apa to Hr Xomw tc5 ZA igov eaTiv. xal em\ iadxi<; laxl TtoXXaTtXdaiov to AE tou TZ xal to EB tou Hr, I'aov 8e to Hr iu AZ, ladxic; apa sgtI TioXXauXdaiov to AE tou TZ xal to EB tou ZA. ladxic 8s UTioxeiTai TtoXXaTtXdaiov to AE tou TZ xal to AB tou EA iadxi<; apa eotI TtoXXaTtXdaiov to EB tou ZA xal to AB tou TA. xal XoiTtov apa to EB Xoitiou tou ZA iadxic; eaTai TtoXXaTtXdaiov, oaaTtXdaiov eaTiv oXov to AB oXou tou TA. 'Eav apa [leye'doz [leye'douci iadxn; fj TtoXXaTtXdaiov, OTtep dcpaipcdev dcpaipcdevToc;, xal to XoiTtov tou XoiTtoO iadxic eaTai TtoXXaTtXdaiov, oaaTtXdaiov eaTi xal to oXov tou oXou - OTtep e5ei 8elc;ai. t In modern notation, this proposition reads ma - m fS = m (a - fS). f'. 'Eav 8uo jieyei&r) 5uo [ieycOtov iadxic fj TtoXXaTtXdaia, xal dcpaipeiJevTa Tivd twv auTWv iadxic fj TtoXXaTtXdaia, xal Ta XoiTia toTc auTolc f]Toi laa eaTiv fj iadxic auT«v TtoX- XaTtXdaia. Auo yap jieyei&r) Ta AB, TA 8uo ^teycdCSv xfiv E, Z Proposition 5* If a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (re- spectively) then the remainder will also be the same mul- tiple of the remainder as that which the whole (is) of the whole (respectively). A E B i 1 1 1 1 1 — i G C F D i — i 1 1 For let the magnitude AB be the same multiple of the magnitude CD that the (part) taken away AE (is) of the (part) taken away CF (respectively). I say that the re- mainder EB will also be the same multiple of the remain- der FD as that which the whole AB (is) of the whole CD (respectively). For as many times as AE is (divisible) by CF, so many times let EB also have been made (divisible) by CG. And since AE and EB are equal multiples of CF and GC (respectively), AE and AB are thus equal multiples of CF and GF (respectively) [Prop. 5.1]. And AE and AB are assumed (to be) equal multiples of CF and CD (respectively). Thus, AB is an equal multiple of each of GF and CD. Thus, GF (is) equal to CD. Let CF have been subtracted from both. Thus, the remainder GC is equal to the remainder FD. And since AE and EB are equal multiples of CF and GC (respectively), and GC (is) equal to DF, AE and EB are thus equal multiples of CF and FD (respectively). And AE and AB are assumed (to be) equal multiples of CF and CD (respectively). Thus, EB and AB are equal multiples of FD and CD (respectively). Thus, the remainder EB will also be the same multiple of the remainder FD as that which the whole AB (is) of the whole CD (respectively) . Thus, if a magnitude is the same multiple of a magni- tude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively) . (Which is) the very thing it was required to show. Proposition 6 f If two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the for- mer magnitudes) are equal multiples of the latter (mag- nitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples 136 ETOIXEIfiN z. ELEMENTS BOOK 5 ladxu; eaxw TtoXXaTtXdaia, xal dcpaipedevxa xd AH, TO xuv auxwv xGv E, Z ladxu; eaxw TtoXXaTtXdaia- Xeyw, oxi xdi Xomd xd HB, 0A xoTg E, Z f]xoi laa eaxiv f] Eadxic; auxcov TtoXXaTtXdaia. A H B i 1 1 1 1 Ei 1 k r © a I 1 1 1 1 1 Z' — ' TEaxco yap Ttpoxepov xo HB iw E l'aov Xeyw, oxi xdi xo 0A t& Z i'aov eaxiv. Keicrda) ydp xw Z iaov xo TK. eTtel iadxi<; eaxl TtoX- XaTtXdaiov xo AH xou E xdi xo T9 xou Z, i'aov 8e xo ^iev HB iG E, xo 8e KT xo Z, ladxi? dpa eaxl TtoXXaTtXdaiov xo AB xou E xdi xo K6 xou Z. ladxu; 8e UTtoxeixai TtoXXaTtXdaiov xo AB xou E xdi xo LA xou Z - ladxi? dpa eaxl TtoXXaTtXdaiov xo K9 xou Z xdi xo TA xou Z. etc! ouv exdxepov xGv K9, TA xou Z tadxu; eaxl TtoXXaTtXdaiov, i'aov dpa eaxl xo K0 iS TA. xoivov dcpr)pr]a , do xo T0 - Xoitiov dpa xo Kr Xomw xo OA laov eaxiv. dXXd xo Z iw Kr eaxiv taov xal xo 6A dpa xw Z Taov eaxiv. waxe ei xo HB xo E laov eaxiv, xal xo OA laov eaxai xw Z. 'Ojioioi; 8r) 5e(?o^.ev, oxi, xav uoXXaTiXdaiov f) xo HB xou E, xoaauxaTiXdaiov eaxai xal xo 6A xou Z. Edv dpa 8uo ^.eyeiDr] 8uo ^eye'dQv ladxu; f) 710X- XaiiXdaia, xal dcpaipeiDevxa xivd xwv auxov ladxic f) 710X- XaiiXdaia, xal xa Xomd xoT<; auxolc f]xoi laa eaxiv f] iadxn; auxCSv TioXXauXdaia- oTiep e8ei SeT^ai. t In modern notation, this proposition reads ma - na = (m - n) a. c Td iaa npbc, xo auxo xov auxov exei Xoyov xal xo auxo izpoq xd laa. TSaxw I'aa ^teye'dr] xd A, B, aXXo 8e xi, o exuxev, ^eyeiDot; xo T- Xeyw, oxi exdxepov xwv A, B 7ip6<; xo T xov auxov e)(ei Xoyov, xal xo T up6<; exdxepov xwv A, B. of them (respectively) . For let two magnitudes AB and CD be equal multi- ples of two magnitudes E and F (respectively) . And let the (parts) taken away (from the former) AG and CH be equal multiples of E and F (respectively) . I say that the remainders GB and HD are also either equal to E and F (respectively), or (are) equal multiples of them. A G B 1 1 1 1 1 Ei ' K C H D 1 1 1 1 1 1 Fi ' For let GB be, first of all, equal to E. I say that HD is also equal to F. For let CK be made equal to F. Since AG and CH are equal multiples of E and F (respectively), and GB (is) equal to E, and KG to F, AB and KH are thus equal multiples of E and F (respectively) [Prop. 5.2]. And AB and CD are assumed (to be) equal multiples of E and F (respectively). Thus, KH and CD are equal multiples of F and F (respectively). Therefore, KH and CD are each equal multiples of F. Thus, KH is equal to CD. Let CH have be taken away from both. Thus, the remainder KG is equal to the remainder HD. But, F is equal to KG. Thus, HD is also equal to F. Hence, if GB is equal to E then HD will also be equal to F. So, similarly, we can show that even if GB is a multi- ple of E then HD will also be the same multiple of F. Thus, if two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively). (Which is) the very thing it was required to show. Proposition 7 Equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ra- tio) to the equal (magnitudes) . Let A and B be equal magnitudes, and C some other random magnitude. I say that A and B each have the 137 ETOIXEIfiN z. ELEMENTS BOOK 5 A 1 1 A 1 1 1 1 1 B 1 1 E 1 1 1 1 1 r> ' Z' ' ' ' EiXf|(p , v)w yap twv jiev A, B iadxu; TtoXXaTiXdaia xd A, E, toD 8e r dXXo, 6 stu)(sv, TtoXXauXdaiov to Z. EtceI ouv ladxic eaxl TtoXXaTtXdaiov to A tou A xal to E tou B, I'aov 8e to A tw B, laov dpa xal to A tw E. dXXo M, 6 ctu/cv, to Z. Ei dpa UTtepexei to A tou Z, UTtepexei xal to E tou Z, xal el Taov, laov, xal si eXaTTov, eXaTTov. xai eaTi Td [lev A, E twv A, B ladxic; TtoXXaitXdaia, to Be: Z tou r dXXo, o eTU^ev, TtoXXaTiXdaiov eotiv dpa w<; to A 7ip6<; to T, outco? to B 7ipo<; to T. Asyw [8/|], oti xal to T 7tp6<; exaTepov twv A, B tov auTov e)(Ei Xoyov. Twv yap auTWv xaTaaxeuaa'devTWv 6|ioiw<; Bel^o^iev, oti laov ecru to A iw E- dXXo 8e ti to Z- ei dpa UTtepexei to Z tou A, UTiepexei xal tou E, xal el laov, I'aov, xal ei eXaTTov, eXaTTov. xai soti to [lev Z tou T TtoXXaTtXdaiov, Ta 8e A, E twv A, B aXXa, a stu^sv, ladxic; TtoXXauXdaia' eaTiv dpa w<; to T Ttp6<; to A, outwc to T npbc, to B. Ta I'aa dpa Kpo<; to auTo tov ai)Tov e^ei Xoyov xal to ai)TO Ttpoc; Ta loot. same ratio to C, and (that) C (has the same ratio) to each of A and B. A i 1 D 1 1 1 1 1 B 1 1 E 1 1 1 1 1 C 1 1 F i 1 1 1 For let the equal multiples D and E have been taken of A and B (respectively), and the other random multiple FofC. Therefore, since D and E are equal multiples of A and B (respectively), and A (is) equal to B, D (is) thus also equal to E. And F (is) different, at random. Thus, if D exceeds F then E also exceeds F, and if (D is) equal (to F then E is also) equal (to F), and if (D is) less (than F then E is also) less (than F) . And D and E are equal multiples of A and f? (respectively), and F another random multiple of C. Thus, as A (is) to C, so B (is) to C [Def. 5.5]. [So] I say that C 1 * also has the same ratio to each of A and B. For, similarly, we can show, by the same construction, that D is equal to E. And F (has) some other (value). Thus, if F exceeds D then it also exceeds E, and if (F is) equal (to D then it is also) equal (to E), and if (F is) less (than D then it is also) less (than E) . And F is a multiple of C, and _D and E other random equal multiples of A and B. Thus, as C (is) to A, so C (is) to B [Def. 5.5]. Thus, equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes). I16piG[Jia. Ex 8r) toutou cpavepov, oti edv (jLEye'Or] Tiva dvdXoyov fl, xal dvdrcaXiv dvdXoyov eaTai. oiiep s8ei 8eT<;ai. t The Greek text has "E", which is obviously a mistake. t In modern notation, this corollary reads that if a : fi : : 7 : 5 then (3 : a Corollary* So (it is) clear, from this, that if some magnitudes are proportional then they will also be proportional inversely. (Which is) the very thing it was required to show. :: 5 : 7. Twv dvlawv jieycdcov to \±ei^ov Ttp6<; to ai)To ^.el^ova Xoyov eyei f\nep to eXaTTov. xal to auTo npbc, to eXaTTov Xoyov sysi. fjjiep 7ipo<; to \leXC,o\. 'EaTW dviaa ^teye'dr] Ta AB, T, xal sotw ^.el^ov to AB, dXXo 8e, o etu)(Ev, to A' Xeyw, otl to AB npbz to A ^.el^ova Xoyov e/et ^itep to T npbq to A, xal to A npbz to T ^el^ova Xoyov ex ei W S P npbc, to AB. Proposition 8 For unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude) . And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater. Let AB and C be unequal magnitudes, and let AB be the greater (of the two), and D another random magni- tude. I say that AB has a greater ratio to D than C (has) to D, and (that) D has a greater ratio to C than (it has) to AB. 138 ETOIXEIfiN z. ELEMENTS BOOK 5 A E B A E B A E B A E B r h K h A h H e H (-) H H K A A N ' 1 1 1 ' N Tkei yap [iei£6v eaxi to AB tou T, xeiai9a> to T laov to BE - to 8rj eXaaaov iSv AE, EB TroXXaTtXaaiaCo^iEvov ecrcoa ttots toO A ^eTi^ov. screw npoxepov to AE sXaTTov toO EB, xal KETioXXaTiXaaida'do to AE, xal eaTW auToO TioXXauXdaiov to ZH \±s%ov ov tou A, xal oaaitXdaiov saTi to ZH toO AE, ToaauTanXdaiov yEyovsTO xal to \±ev H9 toO EB to Se K tou T- xal dXVjcp'da) toO A SmXdaiov jiev to A, TpiTiXdaiov 5e to M, xal Ec;fjc; svl TtXaov, av to Xa^pavo^tevov TioXXauXdaiov [lev yev/]Tai toO A, uptOTC*; 8e ^b£ov tou K. eiXricp'dw, xal eoto to N TSTpauXdaiov jiev tou A, upwTwc; 8e ^ta^ov tou K. 'Etc! ouv to K tou N TtpwTCOc; sotIv eXaTTov, to K dpa tou M oux eaTiv eXaTTov. xal euel iadxic; sotI TtoXXairXdaiov to ZH tou AE xal to H0 tou EB, iadxic; dpa taxi 710X- XaTtXdaiov to ZH tou AE xal to Z8 tou AB. iadxic; 8£ eaTi TtoXXanXdaiov to ZH tou AE xal to K tou T- iadxic; dpa taxi TioXXanXdaiov to ZO tou AB xal to K tou T. xa Z0, K dpa twv AB, T iadxic; eoxl TtoXXaTtXdaia. udXiv, end iadxic; eotI TtoXXauXdaiov to HO tou EB xal to K tou T, taov 8e to EB x& T, Taov dpa xal to H0 tw K. to Se K tou M oux eaTiv eXaTTov ou8' dpa to H0 tou M eXaTTov eaTiv. ^.eI^ov Be to ZH tou A' oXov dpa to Z0 auvaji- cpoTepwv twv A, M [Lel^ov eaTiv. dXXa auva^tcpoTcpa Ta A, M tw N eaTiv I'aa, £7iei8r]Ti£p to M tou A TpiTtXdaiov eaTiv, auva^icpoTspa 8s Ta M, A tou A taxi TCTpauXdaia, ecrci 8e xal to N tou A TSTpajiXdaiov auvajicpoTcpa dpa Ta M, A to N laa eaTiv. dXXa to ZO tgSv M, A \ieiZ,6v eaTiv to Z0 dpa tou N bizepiyev to 8e K tou N oux UTtepe/ei. xal taxi Ta (lev Z0, K t£>v AB, T iadxic; TioXXaTtXdaia, to 8e N tou A dXXo, o £TU)(ev, TioXXaTtXdaiov to AB dpa Tipoc; to A jiei^ova Xoyov e)(ei f\Ksp to T Tipoc; to A. Aeyco 8r|, oti xal to A Tipoc; to T ^lei^ova Xoyov s^si f\mp to A Tipoc; to AB. Twv yap auTQv xaTaaxeuaa'devTWv ojioiwc; Seic^o^iev, oti to ^tev N tou K (mepexei, to Se N tou Z9 oux UTiepexei. xai eaxi to fiev N tou A TioXXaTiXdaiov, Ta 8e Z6, K twv AB, r dXXa, a ctuxev, iadxic; TioXXaTiXdaia- to A dpa Tipoc; to T ^iei£ova Xoyov exei f]7iep to A Tipoc; to AB. AXXa Br] to AE tou EB ^la^ov eoto. to 8r) eXaTTov to EB TtoXXaTiXaaia£6|i£vov eaTai tcots tou A [ie%ov. ne- K ^ D > L ^ K ^ D > L ^ N i 1 1 1 1 N 1 1 1 1 1 For since AB is greater than C, let be made equal to C. So, the lesser of AE and EB, being multiplied, will sometimes be greater than D [Def. 5.4]. First of all, let AE be less than EB, and let AE have been multiplied, and let FG be a multiple of it which (is) greater than D. And as many times as FG is (divisible) by AE, so many times let GH also have become (divisible) by EB, and K by C. And let the double multiple L of D have been taken, and the triple multiple M, and several more, (each increasing) in order by one, until the (multiple) taken becomes the first multiple of D (which is) greater than K. Let it have been taken, and let it also be the quadruple multiple N of D — the first (multiple) greater than K. Therefore, since K is less than A^ first, K is thus not less than M. And since FG and GH are equal multi- ples of AE and EB (respectively), FG and FH are thus equal multiples of AE and AB (respectively) [Prop. 5.1]. And FG and K are equal multiples of AE and C (re- spectively) . Thus, FH and K are equal multiples of AB and C (respectively). Thus, FH, K are equal multiples of AB, C. Again, since GH and K are equal multiples of EB and C, and EB (is) equal to C, GH (is) thus also equal to K. And K is not less than M. Thus, GH not less than M either. And FG (is) greater than D. Thus, the whole of FH is greater than D and M (added) together. But, D and M (added) together is equal to N, inasmuch as M is three times D, and M and D (added) together is four times D, and N is also four times D. Thus, M and D (added) together is equal to N. But, FH is greater than M and D. Thus, FH exceeds N. And K does not exceed N. And FH, K are equal multiples of AB, C, and N another random multiple of D. Thus, AB has a greater ratio to I? than C (has) to I? [Def. 5.7]. So, I say that D also has a greater ratio to C than D (has) to AB. For, similarly, by the same construction, we can show that N exceeds K, and N does not exceed FH. And A^ is a multiple of D, and FH, K other random equal multiples of AB, C (respectively) . Thus, D has a greater 139 ETOIXEIfiN z. ELEMENTS BOOK 5 TtoXXarcXaaida'dM, xal eaxw to HO TtoXXaTtXdaiov \xz\ tou EB, jiei^ov 8e tou A- xal oaaitXdaiov tou to HO tou EB, ToaauTarcXdaiov yeyoveTCO xal to [lev ZH tou AE, to Se K tou r. b\ioicdci 8r) Bei^o^iev, oxi Ta ZO, K twv AB, T ladxic; eaTi TioXXaTtXdaia- xal eiXricp'dw 6[io(w<; to N TtoXXanXdaiov ^terv tou A, TipwT«<; 5e fielCov tou ZH- uaie TidXiv to ZH tou M oux eaTiv sXaaaov. ^ieI^ov 8e to HO tou A - 6Xov apa to ZO twv A, M, toutegti. tou N, hizepe-^ei. to 8s K tou N oux uneps/si, en:£i8r]Tiep xal to ZH ^el^ov ov tou HO, toutsoti tou K, tou N oux UTtEpe/ei. xal (i>aauT(x><; xaTaxoXou-douvTsg tou; £Tidv« Tiepaivojisv ttjv dnoSsi^iv. Tcov apa dviaiov ^teyer'dwv to jisI^ov Tipoc; to auTO (lei^ova Xoyov E)(£i fjjiEp to eXaTTov xal to ain6 Ttpoc; to eXaTTOv [idi^ova Xoyov s)(ei rjnep iipo? to ^ei£ov oitep sSei 8eT^ai. ft'. Ta Tipog to auTO tov auTov e)(ovTa Xoyov laa dXXrjXou; sotiv xal 7ipo<; a to aura tov auTov e)(ei Xoyov, exelva laa SGTIV. A' 1 Bi 1 r> 1 'E)(£tw yap exaTEpov xwv A, B rcpoc; to T tov auTov Xoyov Xeyco, oti Taov ecru, to A tw B. EE yap \ir\, oux av exaTepov twv A, B 7tp6c to T tov auTov elx £ Xoyov £)(£i 8e- 1'aov apa eotI to A iu B. 'E)(£tcl) 8r) udXiv to r 7ip6<; exaTepov twv A, B tov auTov Xoyov Xsyco, oti laov eaTi to A tw B. El yap oux av to T npoc, exaTspov xSv A, B tov auTov el)(£ Xoyov £)(ei 8e- i'aov apa eotI to A iu B. Ta apa 7tp6<; to auTO tov auTov £)(ovTa Xoyov laa dXXfjXou; law xal 7tp6<; a to auTO tov auTov £)( £l Xoyov, exelva laa eaTiv oitep eBei SeTc;ai. l . Tcov npoc, to auTO Xoyov £)(6vt«v to jiei^ova Xoyov e/ov exelvo ^tsT^ov caw 7tp6<; o 8e to ai)To ^ici^ova Xoyov ratio to C than D (has) to ^4_B [Def. 5.5]. And so let AE be greater than EB. So, the lesser, EB, being multiplied, will sometimes be greater than D. Let it have been multiplied, and let GH be a multiple of EB (which is) greater than D. And as many times as GH is (divisible) by EB, so many times let FG also have become (divisible) by AE, and K by C. So, similarly (to the above), we can show that FH and K are equal multiples of AB and C (respectively) . And, similarly (to the above), let the multiple of D, (which is) the first (multiple) greater than FG, have been taken. So, FG is again not less than M. And GH (is) greater than D. Thus, the whole of FH exceeds D and M, that is to say N. And K does not exceed N, inasmuch as FG, which (is) greater than GH — that is to say, K — also does not exceed N. And, following the above (arguments), we (can) complete the proof in the same manner. Thus, for unequal magnitudes, the greater (magni- tude) has a greater ratio than the lesser to the same (mag- nitude) . And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater. (Which is) the very thing it was required to show. Proposition 9 (Magnitudes) having the same ratio to the same (magnitude) are equal to one another. And those (mag- nitudes) to which the same (magnitude) has the same ratio are equal. A i 1 B 1 C' 1 For let A and B each have the same ratio to C. I say that A is equal to B. For if not, A and B would not each have the same ratio to C [Prop. 5.8]. But they do. Thus, A is equal to B. So, again, let C have the same ratio to each of A and B. I say that A is equal to B. For if not, C would not have the same ratio to each of A and B [Prop. 5.8]. But it does. Thus, A is equal to B. Thus, (magnitudes) having the same ratio to the same (magnitude) are equal to one another. And those (magni- tudes) to which the same (magnitude) has the same ratio are equal. (Which is) the very thing it was required to show. Proposition 10 For (magnitudes) having a ratio to the same (mag- nitude), that (magnitude which) has the greater ratio is 140 ETOIXEIfiN z. ELEMENTS BOOK 5 £/£i, exelvo eXaTTov eaTiv. A' 1 Bi 1 r< 1 'E^etco yap to A Trpoc; to V jiei^ova Xoyov fjicep to B iipot; to r- Xeyto, oti [icTCov eaTi to A tou B. Et yap \jx\, fjxoi Taov eaTi to A i« B ij eXaaaov. iaov \iev ouv oux cotI to A tw B' exaTepov yap av twv A, B 7ip6<; to r tov auTov eT/s Xoyov. oux exei 8e' oux apa iaov eaTi to A tu B. ou8e [ir]-v eXaaaov ecra to A tou B- to A yap av Tipoc, to T eXdaaova Xoyov sl/ev f]Trsp to B Trpoc. to T. oux e^ei Se' oux apa eXaaaov 6cm to A tou B. eSsi^i!)/] 8s ou5e loov jieTCov apa eaTi to A tou B. 'E^stw 8rj TidXiv to T npbc, to B jiei^ova Xoyov f]7rep to r Ttpoc. to A' Xeyw, oti eXaaaov coti to B tou A. El yap y.r\, f]Toi iaov eaTiv f\ y.e%ov. Iaov ^tev ouv oux eaTi to B t« A- to T yap dv rcpoc. exaTepov t«v A, B tov auTov eT)(£ Xoyov. oux e^ei 56' °'- ,x "P a ' aov £ax ' 1 T ° A tu B. ouSe u/]v ^el£6v eaTi to B tou A- to T yap av TTpoc, to B eXdaaova Xoyov eTxev r\mp npbz to A. oux ex £l 5e' oux apa pie'iCov eaTi to B tou A. eBeix^r) Se, oti ou5e iaov eXaTTov apa eaTi to B tou A. Twv apa irpog to ai)To Xoyov cxovtwv to [ici^ova Xoyov exov (jieTCov eaTiv xal Trpoc, o to auTO jiei^ova Xoyov exei, exeTvo eXaTTov eaTiv oirep eSei SeT^ai. ia'. Oi tw auTW Xoyco ol auTol xal dXXiqXoic, eialv oi auToi. A i — i r ' ' E' ' B' — i A 1 Z 1 H 1 1 1 ® 1 1 1 K 1 1 1 Ai — i — i — i M 1 1 1 N 1 1 1 1 'EaTwaav yap (be; ^iev to A 7rpo<; to B, outo<; to T npbc, to A, (b<; 6e to r Ttpog to A, outwi; to E 7ipo<; to Z - Xeya>, oti eaTiv (bg to A 7ipo<; to B, outoc to E irpoc; to Z. EiXrjcp'dco yap tuv A, T, E Eadxi? iroXXaTiXdaia Ta H, 0, K, t(5v 5e B, A, Z aXXa, a stuxsv, Eadxi? TroXXaTiXdaia Ta A, M, N. Kal ettel eaTiv (be, to A irpoc to B, outgjc; to T npoc to A, xal eiXrjTTTai tSv [lev A, T ladxig noXXaTiXdaia Ta H, 0, twv 8e B, A aXXa, a eTUxev, ladxic, TioXXairXdaia Ta A, M, ei apa uirepexei to H tou A, UTiepexei xal to tou M, xal ei iaov eaTiv, Iaov, xal ei eXXemei, eXXemei. TidXiv, cttci eaTiv (the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser. A i 1 B 1 C' 1 For let A have a greater ratio to C than B (has) to C. I say that A is greater than B. For if not, A is surely either equal to or less than B. In fact, A is not equal to B. For (then) A and B would each have the same ratio to C [Prop. 5.7]. But they do not. Thus, A is not equal to B. Neither, indeed, is A less than B. For (then) A would have a lesser ratio to C than B (has) to C [Prop. 5.8]. But it does not. Thus, A is not less than B. And it was shown not (to be) equal either. Thus, A is greater than B. So, again, let C have a greater ratio to B than C (has) to A. I say that B is less than A. For if not, (it is) surely either equal or greater. In fact, B is not equal to A. For (then) C would have the same ratio to each of A and B [Prop. 5.7]. But it does not. Thus, A is not equal to B. Neither, indeed, is B greater than A. For (then) C would have a lesser ratio to B than (it has) to A [Prop. 5.8]. But it does not. Thus, B is not greater than A. And it was shown that (it is) not equal (to A) either. Thus, B is less than A. Thus, for (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser. (Which is) the very thing it was required to show. Proposition lit (Ratios which are) the same with the same ratio are also the same with one another. A 1 C' 1 Ei 1 B — i D 1 F' 1 Gi 1 1 Hi 1 1 Ki 1 1 Li — i — i — i M 1 1 1 N 1 1 1 1 For let it be that as A (is) to B, so C (is) to D, and as C (is) to D, so E (is) to F. I say that as A is to B, so E (is) to F. For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively). And since as A is to B, so C (is) to D, and the equal multiples G and H have been taken of A and C (respec- tively), and the other random equal multiples L and M of B and D (respectively), thus if G exceeds L then H also exceeds M, and if (G is) equal (to L then H is also) 141 ETOLXEIftN z. ELEMENTS BOOK 5 (be; to r Ttpoc; to A, outcoc; to E Ttpoc; to Z, xdi dXrjUTai tcov r, E loaxic, TtoXXaTtXdaia Ta 0, K, tcov 5e A, Z dXXa, a etu^ev, iadxic; TtoXXaTtXdaia Ta M, N, el dpa UTtepexei TO 9 tou M, uitepexei xal to K tou N, xal si laov, l'aov, xal el eXXaTov, eXaTTov. dXXa ei UTtepeTxe to O tou M, UTtepeTxe xal to H tou A, xal ei I'aov, I'aov, xal ei eXaTTov, eXaTTov cbaTe xal ei UTtepexei to H tou A, UTtepexei xal to K tou N, xal ei laov, I'aov, xal ei eXaTTov, eXaTTov. xai coti Ta uev H, K tcov A, E iadxic; TtoXXaTtXdaia, Ta Be A, N iSv B, Z dXXa, a CTU)(ev, iadxic; TtoXXaTtXdaia' cotiv dpa cbc; to A Ttpoc; to B, outcoc; to E Ttpoc; to Z. Oi dpa tco aincp Xoycp oi auToi xal dXXr|Xoic; eiaiv oi auToi- oTtep e8ei SeT^ai. equal (to M), and if (G is) less (than L then H is also) less (than M) [Def. 5.5]. Again, since as G is to D, so E (is) to F, and the equal multiples H and K have been taken of G and E (respectively), and the other random equal multiples M and N of D and F (respectively), thus if H exceeds M then K also exceeds N, and if (H is) equal (to M then K is also) equal (to AO, and if (H is) less (than M then K is also) less (than N) [Def. 5.5]. But (we saw that) if H was exceeding M then G was also ex- ceeding L, and if (H was) equal (to M then G was also) equal (to L), and if (H was) less (than M then G was also) less (than L). And, hence, if G exceeds L then K also exceeds AT, and if (G is) equal (to L then AT is also) equal (to AO, and if (G is) less (than L then K is also) less (than N). And G and K are equal multiples of A and (respectively), and L and A^ other random equal multiples of B and F (respectively). Thus, as A is to B, so E (is) to F [Def. 5.5]. Thus, (ratios which are) the same with the same ratio are also the same with one another. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : (3 :: 7 : 8 and 7 : S :: e : C then a : /? :: e : C- 'Eav fj oTiooaouv [ieye'Or) dvaXoyov, eoTai cbc; ev tGv f)youuevcov npbc, ev tGv enotievwv, outgk anavTa Ta fjyouueva Ttpoc; aTtavTa Ta CTto^ieva. Ai ' r> ' ei — ' B' 1 ' Z' 1 H ©1 1 Mi 1 K' ' N' ' 'EaTwaav oTtoaaouv [icye'dr) dvaXoyov Ta A, B, T, A, E, Z, cbc; to A Ttpoc to B, outcoc; to T Ttpoc; to A, xai to E Ttpoc; to Z- Xeyw, oti cotIv wc; to A Ttpoc; to B, outioc; Ta A, r, E Ttpoc; Ta B, A, Z. EiXiqcp-dw yap tGv \iev A, F, E iadxic; TtoXXaTtXdaia Ta H, 0, K, t£5v 8e B, A, Z dXXa, a ctu^cv, iadxic; TtoXXaTtXdaia Td A, M, N. Kai CTtei eaTiv cbc; to A Ttpoc; to B, outcoc; to T Ttpoc; to A, xal to E Ttpoc; to Z, xal el'XrjTtTai tcov (lev A, T, E iadxic; TtoXXaTtXdaia Ta H, Q, K tcov 5e B, A, Z dXXa, d eTU/ev, iadxic; TtoXXaTtXdaia Ta A, M, N, ei dpa UTtepexei to H tou A, UTtepexei xai to 8 tou M, xal to K tou N, xal ei laov, i'aov, xal ei eXaTTov, eXaTTov. cootc xai ei UTtepexei to H tou A, Proposition 12+ If there are any number of magnitudes whatsoever (which are) proportional then as one of the leading (mag- nitudes is) to one of the following, so will all of the lead- ing (magnitudes) be to all of the following. A' 1 C' ' E' ' B 1 D 1 F 1 Hi 1 Mi 1 Ki 1 Ni 1 Let there be any number of magnitudes whatsoever, A, B, C, D, E, F, (which are) proportional, (so that) as A (is) to B, so G (is) to D, and E to F. I say that as A is to B, so A, C, E (are) to B, D, F. For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively). And since as A is to B, so G (is) to D, and E to F, and the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively), thus if G exceeds L then H also exceeds M, and K (exceeds) N, and if (G is) equal (to L then H is also) equal (to M, and K to N), A 1 G' 1 Li 142 ETOIXEIfiN z. ELEMENTS BOOK 5 UTtepexei xod xd H, 0, K iwv A, M, N, xal ei iaov, laa, xal ei eXaxxov, eXaxxova. xai eaxi xo \itv H xal xd H, 0, K xoO A xal xfiv A, T, E ladxic TtoXXanXdaia, ETX£i8r]TXEp edv fj oTioaaouv [icyc'dr) oTtoawvouv [leycdfiv iowv xo TtXrj'doc exaaxov exdaxou ladxic TioXXanXdaiov, oaaitXdaiov eaxiv ev iSv ^teye-duv evoc, xoaauxauXdaia eaxai xal xd Ttdvxa iSv Tidvxcov. Bid xd auxd 8f] xal xo A xal xd A, M, N xoO B xal xfiv B, A, Z ladxu; eaxi TtoXXaitXdaia- eaxiv apa (be xo A npbc, xo B, ouxw<; xd A, T, E Tipoc xd B, A, Z. 'Edv apa fj oTtoaaouv [icyc'dr] dvdXoyov, eaxai 6c ev xwv f]you^iv«v npbz ev xwv enofievuv, ouxgk auavxa xd f]yo6tX£va 7tp6<; arcavxa xd en6\ie\)&- ojiep eBei BeT^ai. and if (G is) less (than L then ii is also) less (than M, and K than AO [Def. 5.5]. And, hence, if G exceeds L then G, if, K also exceed L, M, N, and if (G is) equal (to L then G, ff, K are also) equal (to L, M, N) and if (G is) less (than L then G, if, if are also) less (than L, M, N). And G and G, if, if are equal multiples of A and A, G, f? (respectively), inasmuch as if there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second) [Prop. 5.1]. So, for the same (reasons), L and L, M, N are also equal multiples of B and B, D, F (respectively). Thus, as A is to B, so A, C, E (are) to B, D, F (respectively). Thus, if there are any number of magnitudes whatso- ever (which are) proportional then as one of the leading (magnitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following. (Which is) the very thing it was required to show. + In modern notation, this proposition reads that if a : a' :: f} :/?':: 7 : 7' etc. then a : a' :: (a + + 7 H ) : (a' + (3' + 7' H ). 'Edv upGxov npbz 8euxepov xov auxov !)(T) Xoyov xal xpixov 7tpo<; xexapxov, xplxov Be 7ip6<; xexapxov ^.sl^ova Xoyov exTi fj Tisjinxov npbz Ixxov, xal upaixov upoc; Beuxepov (lei^ova Xoyov s^ei fj ttejitixov npbz exxov. A ' r ' Ei ' B — ' A' ' Z ' Mi 1 iH' ' © ' ' N 1 — 1 — 1 — 1 K 1 1 1 1 A 1 1 1 1 ripGxov yap xo A Tipoc; Beuxepov xo B xov auxov e)(exw Xoyov xal xpixov xo T npbq xfxapxov xo A, xpixov Be xo T 7ip6<; xexapxov xo A ^ie[£ova Xoyov exexw r\ TieuTixov xo E Tipoc; exxov xo Z. Xeyco, 0x1. xal Tipoxov xo A 7ip6<; Beuxepov xo B [iciCova Xoyov e^ei fjuep TtejiTtxov xo E Tipoc; exxov xo Z. 'EticI yap eaxi xivd xov ^xev T, E ladxic TioXXauXdaia, xCSv Be A, Z dXXa, a exu)(ev, ladxic noXXaTiXdoia, xal xo (iev xoO r TioXXaTiXdoiov xou xou A TtoXXauXaabu bnspiyzi, xo Be xou E TioXXajcXdaiov xou xou Z TioXXaicXaaiou ou)( ujcepexei, EiXfjqj'dco, xal eaxw xwv ^.ev T, E ladxic tcoX- XaicXdaia xd H, 0, xwv Be A, Z dXXa, a exu)(ev, ladxic TioXXauXdaia xd K, A, «axe xo [lev H xou K U7iepe)(eiv, xo Be xou A \±r\ uuepexeiv xal oaaTiXdaiov (lev eaxi xo H xou T, xoaauxanXdaiov eaxco xal xo M xou A, oaauXdaiov Be xo K xou A, xoaauxanXdaiov eaxw xal xo N xou B. Proposition 13 f If a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the third (magnitude) has a greater ratio to the fourth than a fifth (has) to a sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth. A' ' C ' E' 1 B 1 1 D 1 F' 1 Mi 1 1 G 1 ' Hi 1 1 Ni 1 1 1 K 1 1 1 Li 1 1 1 For let a first (magnitude) A have the same ratio to a second B that a third G (has) to a fourth D, and let the third (magnitude) G have a greater ratio to the fourth D than a fifth E (has) to a sixth F. I say that the first (magnitude) A will also have a greater ratio to the second B than the fifth E (has) to the sixth F. For since there are some equal multiples of G and E, and other random equal multiples of D and F, (for which) the multiple of G exceeds the (multiple) of D, and the multiple of E does not exceed the multiple of F [Def. 5.7], let them have been taken. And let G and H be equal multiples of G and E (respectively), and K and L other random equal multiples of D and F (respectively), such that G exceeds K, but H does not exceed L. And as many times as G is (divisible) by G, so many times let M be (divisible) by A. And as many times as K (is divisible) 143 ETOIXEIfiN z. ELEMENTS BOOK 5 Kal ETtei eaxiv w<; to A Ttpoc; to B, ouxco? to Y Ttpoc; to A, xal eiXrjTtxai xfiiv jjiev A, Y iadxic; TtoXXaTtXdaia xd M, H, iSv 8e B, A aXXa, a exuxev, iadxic; TtoXXaTtXdaia xd N, K, el dpa UTteps/si to M xou N, UTtepexei xal xo H xou K, xod el laov, Taov, xal ei eXaxxov, eXXaxov. Cmepexei Be xo H xou K- UTtepexei dpa xal xo M xou N. xo Se 6 xou A oux UTtepexei 1 xai eaxi xd [iev M, O xaiv A, E iadxic; TtoXXaTtXdaia, xd 6e N, A xfiv B, Z aXXa, a exuxev, Iadxic; TtoXXaTtXdaia- xo dpa A Ttpoc; xo B ^tei^ova Xoyov exei f]nep xo E Ttpoc; xo Z. 'Eav dpa upwxov Ttpoc; Beuxepov xov auxov ex?] Xoyov xal xpixov Ttpoc; xexapxov, xpixov 8e Ttpoc; xexapxov ^tei^ova Xoyov exT) ^ Tte^tTtxov Ttpoc; exxov, xal TtpGxov Ttpoc; 8euxepov ^tei^ova Xoyov ei;ei rj Tte^titxov Ttpoc; exxov oitep e8ei SeTc;ai. by D, so many times let N be (divisible) by B. And since as ^4 is to B, so C (is) to D, and the equal multiples M and G have been taken of A and G (respec- tively), and the other random equal multiples N and K of B and D (respectively), thus if M exceeds N then G exceeds K, and if (M is) equal (to N then G is also) equal (to K), and if (M is) less (than N then G is also) less (than K) [Def. 5.5]. And G exceeds K. Thus, M also exceeds N. And i7 does not exceeds L. And M and H are equal multiples of A and _E (respectively), and N and L other random equal multiples of B and F (respec- tively). Thus, A has a greater ratio to B than (has) to F [Def. 5.7]. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and a third (magni- tude) has a greater ratio to a fourth than a fifth (has) to a sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : (3 :: 7 : 8 and 7 : 8 > e : f then a : (3 > e : (. i5'. 'Eav TtpcSxov Ttpoc; Seuxepov xov auxov exT) Xoyov xal xpixov Ttpoc; xexapxov, xo Se TtpGxov xou xpixou ^icTCov fj, xal xo Seuxepov xou xexdpxou jieT^ov eaxai, xdv Taov, '(gov, xdv eXaxxov, eXaxxov. Ai 1 ri 1 Bi 1 A' ' IlpGxov yap TO A Ttpoc; Seuxepov xo B auxov cxcxm Xoyov xal xpixov xo Y Ttpoc; xexapxov xo A, [leTC^ov Se eaxw xo A xou r- Xeyw, oxi xal xo B xou A fjieT^ov eaxiv. 'Eitel yap xo A xou Y \j.e%6\i eaxiv, dXXo Se, o exuxev, [^eyedoc;] xo B, xo A dpa Ttpoc; xo B ^tei^ova Xoyov exei rjitep xo T Ttpoc; xo B. (be; Se xo A Ttpoc; xo B, ouxwc; xo T Ttpoc; xo A- xal xo T dpa Ttpoc xo A ^eii^ova Xoyov exei f]Ttep xo r Ttpoc; xo B. Ttpoc; o Se xo auxo (jiei^ova Xoyov exei, exelvo eXaaaov eaxiv eXaaaov dpa xo A xou B- waxe ^el£6v eaxi xo B xou A. "O^oicoc; 8rj 8eTi;o^ev, 6x1 xdv taov f) xo A xw Y, Taov eaxai xal xo B iu A, xdv eXaaaov fj xo A xou T, eXaaaov eaxai xal xo B xou A. 'Eav dpa itpwxov Ttpoc; Seuxepov xov auxov exT) Xoyov xal xpixov Ttpoc; xexapxov, xo 8e Ttpwxov xou xpixou y.e%o\> fj, xal xo 8euxepov xou xexdpxou \ie%o\> eaxai, xdv Taov, Taov, xdv eXaxxov, eXaxxov oitep e8ei 8ele;ai. Proposition 14 f If a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the first (magnitude) is greater than the third, then the second will also be greater than the fourth. And if (the first magnitude is) equal (to the third then the second will also be) equal (to the fourth) . And if (the first magnitude is) less (than the third then the second will also be) less (than the fourth) . A 1 1 C' 1 B 1 1 D 1 For let a first (magnitude) A have the same ratio to a second B that a third G (has) to a fourth D. And let A be greater than G. I say that B is also greater than D. For since A is greater than G, and B (is) another ran- dom [magnitude], A thus has a greater ratio to B than G (has) to B [Prop. 5.8]. And as A (is) to B, so G (is) to D. Thus, G also has a greater ratio to D than G (has) to B. And that (magnitude) to which the same (magnitude) has a greater ratio is the lesser [Prop. 5.10]. Thus, D (is) less than B. Hence, B is greater than D. So, similarly, we can show that even if A is equal to G then B will also be equal to D, and even if A is less than G then B will also be less than D. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the first (mag- nitude) is greater than the third, then the second will also be greater than the fourth. And if (the first magnitude is) 144 ETOIXEIfiN z. ELEMENTS BOOK 5 equal (to the third then the second will also be) equal (to the fourth) . And if (the first magnitude is) less (than the third then the second will also be) less (than the fourth) . (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : j3 :: 7 : 8 then a = 7 as f3 = <5. IE . Proposition 15 f Td H£pT] xolc; Gaauxioc; TtoXXauXaaioic; xov aCxov ttyzi Xoyov XTjcpiSevxa xaxdXXr]Xa. A H B 1 1 1 1 j/ 1 1 1 A 1— K — 1 — A — 1 — E —1 2' ' "Eaxw yap ladxic; TtoXXauXdaiov to AB xou T xal to AE toO Z- Xey«, oxi laxlv cbc xo T Ttpoc; xo Z, ouxwc; xo AB 7ip6<; xo AE. 'End yap ladxic; eaxl KoXXairXdoiov xo AB xoO T xal xo AE xou Z, 00a apa eaxlv ev xG AB ^.eyei!)/] laa xG T, xoaaOxa xal ev xG AE Taa xG Z. 5ir]pr]OTf)w xo fjiev AB slg xa xG T I'oa xd AH, H6, 6B, xo 8e AE zlc. xd xG Z Taa xa AK, KA, AE' eaxai 8r| I'aov xo TtXrj'doc; xGv AH, H6, 6B xG TrXrydei xGv AK, KA, AE. xal end Taa eoxl xa AH, H6, 9B dXX^Xoic, eoxi Bs xal xd AK, KA, AE I'oa dXXiqXou;, saxiv apa Gc; xo AH npbc, xo AK, ouxwc; xo H0 Tipog xo KA, xal xo &B npbz xo AE. soxai apa xal Gg §v xGv f)you|jievMv Ttpoc; ev xGv euo^ievwv, ouxwc drcavxa xd f)you[ieva upoc duavxa xd snojieva' eaxiv apa G<; xo AH npbc, xo AK, ouxgk xo AB npbc, xo AE. I'oov 8e xo jisv AH xG T, xo 8s AK xG Z- eoxiv apa Gc; xo T npbc, xo Z ouxcoc xo AB iipoc; xo AE. Td apa [jtspr] xoTg Gaauxwc; TioXXanXaoioic; xov auxov SX^I Xoyov XricpiSevxa xaxdXXrjXa- oiiep eBei BeT^ai. Parts have the same ratio as similar multiples, taken in corresponding order. A G H B 1 1 1 1 Q 1 1 D 1— K E —1 For let AB and DE be equal multiples of C and F (respectively) . I say that as C is to F, so AB (is) to DE. For since AB and DE are equal multiples of C and F (respectively), thus as many magnitudes as there are in AB equal to C, so many (are there) also in DE equal to F. Let AB have been divided into (magnitudes) AG, GH, HB, equal to C, and DE into (magnitudes) DK, KL, LE, equal to F. So, the number of (magnitudes) AG, GH, HB will equal the number of (magnitudes) DK, KL, LE. And since AG, GH, HB are equal to one another, and DK, KL, LE are also equal to one another, thus as AG is to DK, so GH (is) to KL, and HB to LE [Prop. 5.7]. And, thus (for proportional magnitudes), as one of the leading (magnitudes) will be to one of the fol- lowing, so all of the leading (magnitudes will be) to all of the following [Prop. 5.12]. Thus, as AG is to DK, so AB (is) to DE. And AG is equal to C, and DK to F. Thus, as C is to F, so AB (is) to DE. Thus, parts have the same ratio as similar multiples, taken in corresponding order. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that a : f3 : : m a : m f3. If'. 'Edv xeoaapa (ieyei9r) dvdXoyov f), xal evaXXac; dvdXoyov eaxai. "Eax« xeaoapa [leye^r] dvdXoyov xd A, B, T, A, G<; xo A Tipoc xo B, ouxox xo T Tipoc; xo A- Xcyw, oxi xal svaXXd^ [dvdXoyov] eaxai, G<; xo A npoc; xo T, ouxwc; xo B Tipoc xo A. ElXr ! ]cp'dw yap xGv jiev A, B ladxic; TtoXXaTtXaaia xd E, Z, xGv 8e T, A aXXa, a sxuxev, ladxic; TioXXajcXdaia xd H, e. Proposition 16 f If four magnitudes are proportional then they will also be proportional alternately. Let A, B, C and D be four proportional magnitudes, (such that) as A (is) to B, so C (is) to D. I say that they will also be [proportional] alternately, (so that) as A (is) to C, so B (is) to D. For let the equal multiples E and F have been taken of A and B (respectively), and the other random equal multiples G and H of C and D (respectively) . 145 ETOIXEIfiN z. ELEMENTS BOOK 5 B h E i- Z^ H 1 Kal cticI iadxic eaxl TtoXXauXdaiov to E xoO A xal to Z toO B, xa Se (JepT) xolc waauxwc TtoXXaTiXaaioic xov aGxov exei Xoyov, eaxiv apa cbc xo A Tipoc xo B, ouxwc xo E Tipoc xo Z. «c 8e xo A Tipoc xo B, ouxwc xo T Tipoc xo A' xal d>c apa xo r Tipoc xo A, ouxwc xo E Tipoc xo Z. TtdXiv, euel xd H, 6 xwv T, A iadxic eaxl TioXXaTtXdaia, eaxiv apa cbc xo T Tipoc xo A, ouxoc xo H Tipoc xo 0. ci>c 8e xo T Tipoc xo A, [ouxtoc] xo E Tipoc xo Z- xal <i>c apa xo E Tipoc xo Z, ouxcoc xo H Tipoc xo 9. eav 8e xeaaapa ^eyc'dr] dvdXoyov fj, xo 8s Ttpwxov xoO xpixou [isT^ov fj, xal xo 8euxepov xou xexdpxou \ieiZov eaxai, xav laov, iaov, xdv eXaxxov, eXaxxov. ei apa UTiepexei xo E xou H, imepexei xal xo Z xou 0, xal ei laov, laov, xal ei eXaxxov, eXaxxov. xal eaxi xd ^.ev E, Z xwv A, B iadxic TioXXaTtXdaia, xa 8e H, 8 xov T, A aXXa, a exuxev, iadxic TioXXaTtXdaia- eaxiv apa cbc xo A Tipoc xo T, ouxwc xo B Tipoc xo A. 'Eav apa xeaaapa ^.eyedr] dvdXoyov fj, xal evaXXa^ dvdXoyov eaxar oxep e8ei SeT^ai. A^ F O H 1 And since E and F are equal multiples of A and B (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as A is to B, so E (is) to F. But as A (is) to B, so C (is) to D. And, thus, as C (is) to D, so _E (is) to F [Prop. 5.11]. Again, since G and H are equal multiples of C and D (respectively), thus as C is to D, so G (is) to Jf [Prop. 5.15]. But as C (is) to D, [so] £ (is) to F. And, thus, as E (is) to F, so G (is) to iJ [Prop. 5.11]. And if four magnitudes are proportional, and the first is greater than the third then the second will also be greater than the fourth, and if (the first is) equal (to the third then the second will also be) equal (to the fourth), and if (the first is) less (than the third then the second will also be) less (than the fourth) [Prop. 5.14]. Thus, if E exceeds G then F also exceeds H, and if (E is) equal (to G then F is also) equal (to H), and if (E is) less (than G then F is also) less (than H) . And E and F are equal multiples of A and B (respectively), and G and H other random equal multiples of C and D (respectively) . Thus, as A is to C, so B (is) to D [Def. 5.5]. Thus, if four magnitudes are proportional then they will also be proportional alternately. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : (3 :: 7 : 8 then a : 7 :: (3 : 8. 'Edv auyxei^ieva jieye-v}/] dvdXoyov fj, xal 8iaipe-devxa dvdXoyov eaxai. A E B r Z A 1 1 1 1 1 — 1 H S K S 1 1 1 1 A M N n 1 1 1 1 TCaxco auyxeifieva [ieye'Or] dvdXoyov xa AB, BE, IA, AZ, cbc to AB Tipoc xo BE, ouxwc to IA Tipoc to AZ- Xeyco, oxi xal Biaipcdevxa dvdXoyov eaxai, wc xo AE Tipoc xo EB, ouxwc to TZ Tipoc to AZ. EiXrjcp'dco yap xwv ^tev AE, EB, TZ, ZA iadxic TioX- XaTtXdaia xa H9, 6K, AM, MN, xwv 8e EB, ZA aXXa, a exuxev, iadxic TioXXaTtXdaia xa KS, Nil. Kal cticI iadxic saxl TioXXaxXdaiov xo H9 xou AE xal xo OK xou EB, iadxic apa eaxl TtoXXaTiXdaiov xo H6 xoO Proposition 17 f If composed magnitudes are proportional then they will also be proportional (when) separarted. A E B C F D 1 1 1 1 1 1 G H K O 1 1 1 1 L MNP 1 1 1 1 Let AB, BE, CD, and DF be composed magnitudes (which are) proportional, (so that) as AB (is) to BE, so CD (is) to DF. I say that they will also be proportional (when) separated, (so that) as AE (is) to EB, so CF (is) to DF. For let the equal multiples GH, HK, LM, and MN have been taken of AE, EB, CF, and FD (respectively), and the other random equal multiples KO and NP of EB and FD (respectively). 146 ETOIXEIfiN z. ELEMENTS BOOK 5 AE xal to HK tou AB. iadxu; Se ecru TtoXXaitXaaiov to H0 tou AE xal to AM tou TZ- iodxn; apa eaTi TtoXXanXdmov to HK tou AB xal to AM toO TZ. TtdXiv, snel iadxic sgtI TtoXXarcXaaiov to AM tou TZ xal to MN tou ZA, iadxu; apa taxi TroXXanXdaiov to AM tou TZ xal to AN tou IA. ladxig Be rjv KoXXanXdoiov to AM tou TZ xal to HK tou AB- iadxu; apa eaTi TioXXanXdoiov to HK tou AB xal to AN tou TA. Ta HK, AN apa t«v AB, IA iadxu; eotI TtoXXanXdaia. ndXiv, inz\ iadxu; eotI TioXXaTtXaaiov to 0K tou EB xal to MN tou ZA, eaTi Be xal to KS tou EB iadxu; TioXXanXdaiov xal to Nn tou ZA, xal auvTcdev to 9S tou EB iadxu; eaxl TtoXXarcXaaiov xal to Mil tou ZA. xal snei eaTiv cbc to AB 7ip6<; to BE, outo<; to TA 7ip6<; to AZ, xal d'Xr]7iTai xwv [lev AB, TA iadxu; TtoXXaTtXdaia Ta HK, AN, t£Sv 8e EB, ZA iadxu; TtoXXaTtXdaia Ta 9S, Mil, ei apa (mepexei TO HK tou 9S, unepexei xal to AN tou Mil, xal ei laov, laov, xal ei sXaTTov, eXaTTOv. uitepexeTO) 8f| t° HK tou OS, xal xoivou dcpaipcdevTOi; tou 9K UTiepexei. apa xal to H6 tou KS. dXXa ei U7iepeT)(e to HK tou 0S uiiepeTxe xal to AN tou Mil' UTiepexei apa xal to AN tou Mil, xal xoivou dcpaipcdevTot; tou MN UTiepexei xal to AM tou Nn- wcrce ei UTiepexei to H8 tou KS, uuepexei xal to AM tou Nn. o^ioicoc; 8r] SeT^o^tev, oxi xdv laov fj to H8 to KS, taov eaTai xal to AM to Nn, xdv eXaTTOv, eXaTTOv. xai eaTi Ta [lev H0, AM twv AE, TZ iadxu; TioXXaicXdaia, Ta 8e KS, Nn twv EB, ZA dXXa, a ctuxcv, iadxic; icoXXaiiXdaia- cotiv apa ok to AE 7ip6<; to EB, outoc to TZ Ttpoc; to ZA. 'Eav apa auyxei^ieva [leye'dr) dvdXoyov fj, xal 8iai- peiDevTa dvdXoyov eaTar oiiep e8ei SeT^ai. And since GH and HK are equal multiples of AE and EB (respectively), GH and GK are thus equal multiples of AE and AB (respectively) [Prop. 5.1]. But GH and LM are equal multiples of AE and CF (respectively). Thus, GK and LM are equal multiples of AB and CF (respectively) . Again, since LM and MN are equal mul- tiples of CF and FD (respectively), LM and LN are thus equal multiples of CF and CD (respectively) [Prop. 5.1]. And LM and GK were equal multiples of CF and AB (respectively). Thus, GK and LN are equal multiples of AB and CD (respectively). Thus, GK, LN are equal multiples of AB, CD. Again, since HK and MN are equal multiples of EB and FD (respectively), and KO and NP are also equal multiples of EB and FD (respec- tively), then, added together, HO and MP are also equal multiples of FP> and FD (respectively) [Prop. 5.2]. And since as AB (is) to BE, so CD (is) to DF, and the equal multiples GK, LN have been taken of AB, CD, and the equal multiples HO, MP of EB, FD, thus if GK exceeds HO then LN also exceeds MP, and if {GK is) equal (to HO then LN is also) equal (to MP), and if (GK is) less (than HO then LN is also) less (than MP) [Def. 5.5]. So let GK exceed HO, and thus, HK being taken away from both, GH exceeds KO. But (we saw that) if GK was exceeding HO then LN was also exceeding MP. Thus, LN also exceeds MP, and, MN being taken away from both, LM also exceeds NP. Hence, if GH exceeds KO then LM also exceeds NP. So, similarly, we can show that even if GH is equal to KO then LM will also be equal to NP, and even if (GH is) less (than KO then LM will also be) less (than NP). And GH, LM are equal multiples of AE, CF, and KO, NP other random equal multiples of EB, FD. Thus, as AE is to EB, so CF (is) to FD [Def. 5.5]. Thus, if composed magnitudes are proportional then they will also be proportional (when) separarted. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that ifa + /3:/3::7 + <5:<5 then a : f3 :: 7 : 8. 'Edv 8ir)p/)[ieva ^eye^r) dvdXoyov fj, xal auvTeiJevTa dvdXoyov eaxai. A E B Proposition 18+ If separated magnitudes are proportional then they will also be proportional (when) composed. A E B r i— z — i — H —i — A — i 'Eotw 8ir]pr]neva [icye-dr] dvdXoyov Ta AE, EB, TZ, ZA, &>c, to AE Ttpoc; to EB, outoc; to TZ Ttpoc; to ZA- Xeyw, oti xal ouvTeiJevTa dvdXoyov eaxai, tbc; to AB npbc, to BE, c h- G — i — D — i Let AE, EB, CF, and FD be separated magnitudes (which are) proportional, (so that) as AE (is) to EB, so CF (is) to FD. I say that they will also be proportional 147 ETOIXEIftN z. ELEMENTS BOOK 5 ouxcoc; to TA Ttpoc; to ZA. EE yap \xr\ eaxiv tlx; to AB Ttpoc; xo BE, ouxcoc; to TA Ttpoc; xo AZ, eaxai cbc; xo AB Ttpoc; xo BE, ouxcoc; xo TA fjxoi Ttpoc; eXaaaov xi xou AZ fj Ttpoc; jiel^ov. 'Eaxco Ttpoxepov Ttpoc; eXaaaov xo AH. xal stxei eaxiv cbc; xo AB Ttpoc; xo BE, ouxcoc; xo TA Ttpoc; xo AH, auyxei^eva (leyc'dr) dvaXoyov eaxiv wore xal Biaipcdevxa dvaXoyov eaxai. eaxiv dpa cbc; xo AE Ttpoc; xo EB, ouxcoc; xo TH Ttpoc; xo HA. UTtoxeixai 5e xal cbc; xo AE Ttpoc; xo EB, ouxcoc; xo rZ Ttpoc; xo ZA. xal cbc; dpa xo TH Ttpoc; xo HA, ouxcoc; xo rZ Ttpoc; xo ZA. jiel^ov 6e xo Ttpcoxov xo TH xou xpixou xou TZ- [iel^ov dpa xal xo 8euxepov xo HA xou xexdpxou xou ZA. dXXd xal eXaxxov oTtep eaxiv dBuvaxov oux dpa eaxiv cbc; xo AB Ttpoc; xo BE, ouxcoc; xo TA Ttpoc; eXaaaov xou ZA. ojioloK 6f| 6eii;o^ev, oxi ouSe Ttpoc; [MCov Ttpoc; auxo dpa. 'Eav dpa Biflprjjieva |ieyei9r) dvaXoyov rj, xal auvxe-devxa dvaXoyov eaxai- onep eSei 8el^ai. (when) composed, (so that) as AB (is) to BE, so CD (is) to FD. For if (it is) not (the case that) as AB is to BE, so CD (is) to FD, then it will surely be (the case that) as AB (is) to BE, so CD is either to some (magnitude) less than DF, or (some magnitude) greater (than DF)} Let it, first of all, be to (some magnitude) less (than DF), (namely) DC And since composed magnitudes are proportional, (so that) as AB is to BE, so CD (is) to DC, they will thus also be proportional (when) separated [Prop. 5.17]. Thus, as AE is to EB, so CG (is) to CD. But it was also assumed that as AE (is) to EB, so CF (is) to FD. Thus, (it is) also (the case that) as CG (is) to GD, so CF (is) to FD [Prop. 5.11]. And the first (magnitude) CG (is) greater than the third CF. Thus, the second (magnitude) GD (is) also greater than the fourth FD [Prop. 5.14]. But (it is) also less. The very thing is impossible. Thus, (it is) not (the case that) as AB is to BE, so CD (is) to less than FD. Similarly, we can show that neither (is it the case) to greater (than FD). Thus, (it is the case) to the same (as FD) . Thus, if separated magnitudes are proportional then they will also be proportional (when) composed. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : (3 :: 7 : <5 then a + /3 : /3 :: 7 + <5 : <5. * Here, Euclid assumes, without proof, that a fourth magnitude proportional to three given magnitudes can always be found. °Edv fj ok oXov Ttpoc oXov, ouxcoc; dcpaipsiJev Ttpoc; dcpai- pe^Ev, xal xo Xoikov Tcpoc; xo Xomov eaxai «<; oXov Tcpoc; oXov. A E B 1 1 1 r z a 1 1 1 'Eaxto yap tbcj oXov xo AB Ttpoc; oXov xo TA, ouxcoc; dcpaipe'dev xo AE Ttpoc; dcpEipeiDev xo TZ' Xeyco, oxi xal Xomov xo EB Ttpoc; Xomov xo ZA eaxai cbc; oXov xo AB Ttpoc; oXov xo TA. 'EtccI yap eaxiv cbc; xo AB Ttpoc; xo TA, ouxcoc; xo AE Ttpoc; xo TZ, xal evaXXdc; cbc; xo BA Ttpoc; xo AE, ouxcoc; xo AT Ttpoc; xo TZ. xal enel auyxeijieva jieye-d/) dvaXoyov eaxiv, xal 8iaipei5evxa dvaXoyov eaxai, cbc; xo BE Ttpoc; xo EA, ouxcoc; xo AZ Ttpoc; xo TZ- xal evaXXdc;, cbc; xo BE Ttpoc; xo AZ, ouxcoc; xo EA Ttpoc; xo ZT. cb<; 8e xo AE Ttpoc; xo TZ, ouxcoc; UTtoxeixai oXov xo AB Ttpoc; oXov xo TA. xal XoiTtov dpa xo EB Ttpoc; Xoitiov xo ZA eaxai cbc; oXov xo AB Ttpog oXov xo TA. 'Eav dpa fj cbc; oXov Ttpoc oXov, ouxcoc; dtpaipeiJev Ttpoc; Proposition 19 f If as the whole is to the whole so the (part) taken away is to the (part) taken away then the remainder to the remainder will also be as the whole (is) to the whole. A E B 1 1 1 C F D 1 1 1 For let the whole AB be to the whole CD as the (part) taken away AE (is) to the (part) taken away CF. I say that the remainder EB to the remainder FD will also be as the whole AB (is) to the whole CD. For since as AB is to CD, so AE (is) to CF, (it is) also (the case), alternately, (that) as BA (is) to AE, so DC (is) to CF [Prop. 5.16]. And since composed magni- tudes are proportional then they will also be proportional (when) separated, (so that) as BE (is) to EA, so DF (is) to CF [Prop. 5.17]. Also, alternately, as BE (is) to DF, so EA (is) to FC [Prop. 5.16]. And it was assumed that as AE (is) to CF, so the whole AB (is) to the whole CD. And, thus, as the remainder EB (is) to the remainder FD, so the whole AB will be to the whole CD. 148 ETOIXEIfiN z. ELEMENTS BOOK 5 dcpaipsiftsv, xal to Xomov Ttpoc to Xoikov saTai <i>c oXov Ttpoc oXov [oTtsp s8si Ssl^ai]. [Kod STtel eSsix'dr) cbc to AB Ttpoc to IA, outwc to EB Ttpoc to ZA, xal svaXXdc" tlx; to AB Ttpoc to BE outgjc to TA Ttpoc to ZA, auyxsi|isva apa ^sysi}/] dvdXoyov screw sSeix'dr) 8s cbc to BA Ttpoc to AE, outioc to Ar Ttpoc to TZ- xai egtiv dvaoTps^avTi]. Thus, if as the whole is to the whole so the (part) taken away is to the (part) taken away then the remain- der to the remainder will also be as the whole (is) to the whole. [(Which is) the very thing it was required to show] [And since it was shown (that) as AB (is) to CD, so EB (is) to FD, (it is) also (the case), alternately, (that) as AB (is) to BE, so CD (is) to FD. Thus, composed magnitudes are proportional. And it was shown (that) as BA (is) to AE, so DC (is) to CF. And (the latter) is converted (from the former).] II6piG[jia. 'Ex 8r] toutou cpavspov, oti sav auyxsijisva [isys'dr) dvdXoyov fj, xal dvacrcpsij'avTi dvdXoyov so"tar OTtsp s8si BsTc;ai. t In modern notation, this proposition reads that if a : (3 :: 7 : 5 then a : * In modern notation, this corollary reads that if a : (3 :: 7 : 8 then a : a Corollary 11 So (it is) clear, from this, that if composed magni- tudes are proportional then they will also be proportional (when) converted. (Which is) the very thing it was re- quired to show. :: a - 7 : /3 - 8. — (3 :: 7 : 7 — 8. X . 'Edv fj Tpia ^sys^r) xot ' dXXa auTolc Xai to TtXrji&oc, auv8uo Xa(iPav6(jieva xal sv tw ai)TO Xoyw, 81' I'aou Bs to IXpWTOV TOU TpiTOU [ISlCoV fj, Xal TO TSTapTOV TOU SXTOU [isiZov scrcai, xav I'oov, I'oov, xav sXaTTov, sXaTTov. A' 1 A 1 1 Bi 1 Ei 1 Ti 1 Z' 1 'EaTW Tpia ^.sysiDr] Ta A, B, T, xal dXXa auTolc laa to TtXrj$oc Ta A, E, Z, auvBuo Xa^pavo^teva sv tw auTS Xoycp, cbc [isv to A Ttpoc to B, outwc to A Ttpoc to E, «c 8s to B Ttpoc to T, outmc to E Ttpoc to Z, 81' laou 8s fisT^ov sotw to A toO T- Xsyio, oti xal to A tou Z [isi^ov iaxai, xav laov, taov, xav sXaTTov, sXaTTov. 'Etcei yap ^isT£6v scm to A toO T, aXXo 8s ti to B, to 8s ^isT£ov Ttpoc to auTO [isiCova Xoyov sxsi fjusp to sXaTTov, to A apa Ttpoc to B (isi^ova Xoyov s^si fjTtsp to T Ttpoc to B. dXX' cbc ^isv to A Ttpoc to B [outoc] to A Ttpoc to E, cbc 8s to T Ttpog to B, dvdnaXiv outgjc to Z Ttpoc to E- xal to A apa Ttpoc to E [isi^ova Xoyov s^si fjTtsp to Z Ttpoc to E. twv Bs Ttpoc to ai)TO Xoyov sxovtcjv to ^.s[£ova Xoyov s/ov [Lei^bv scmv. [leiZoM apa to A toO Z. o^tolwc 8r] 8s[^o[jisv, oti xav I'aov fj to A tw T, urov saTai xal to A t« Z, xav Proposition 20 f If there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth). And if (the first is) less (than the third then the fourth will also be) less (than the sixth) . A 1 ' D' 1 B 1 1 E 1 1 Ci 1 F 1 1 Let A, B, and C be three magnitudes, and D, E, F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two, (so that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F. And let A be greater than C, via equality. I say that D will also be greater than F. And if (A is) equal (to C then D will also be) equal (to F). And if (A is) less (than C then D will also be) less (than F). For since A is greater than C, and B some other (mag- nitude), and the greater (magnitude) has a greater ratio than the lesser to the same (magnitude) [Prop. 5.8], A thus has a greater ratio to B than C (has) to B. But as A (is) to B, [so] D (is) to E. And, inversely, as C (is) to B, so F (is) to E [Prop. 5.7 corr.]. Thus, D also has a greater ratio to E than F (has) to E [Prop. 5.13]. And for (mag- 149 ETOIXEIfiN z. ELEMENTS BOOK 5 eXaxxov, eXaxxov. 'Eav apa fj xpia (ieye'dr) xal aXXa auxolc; laa xo TtXrji&oc;, auvSuo Xa^pav6|ieva xal ev xw auxw Xoyw, 81° 1'aou 8e xo npwxov xou xpixou [isTCov fj, xal xo xexapxov xou exxou jieT£ov eaxai, xav iaov, iaov, xav eXaxxov, eXaxxov oTtep £8ei 8eTc;ai. t In modern notation, this proposition reads that if a : (3 :: 8 : e and j3 : XOL. 'Eav fj xpia jieye'dr] xal aXXa auxolc; laa xo TtXrjT9oc; auv8uo Xajj.f3av6jj.sva xal ev xS auxfi Xoyw, fj 8e xexa- payjjevrj auxfiiv rj dvaXoyia, 81' 1'aou 8e xo npSxov xou xpixou y.siZ,ov fj, xal xo xexapxov xou exxou (iel^ov eaxai, xav laov, iaov, xav eXaxxov, eXaxxov. A i 1 A i 1 B' 1 E' 1 ri 1 z> ' 'Eaxw xpia jxeye'dr] xa A, B, T xal aXXa auxolc; Xou xo TiXfj'dot; xa A, E, Z, auv8uo Xajjpavojjeva xal ev xQ auxw Xoyco, eax« 8e xexapaYjjevr) auxfiv f\ dvaXoyia, w<; jxsv xo A npo? xo B, ouxdx xo E npo? xo Z, cbc Se xo B Kpot; xo T, ouxw<; xo A 7ip6<; xo E, 81' laou Ss xo A xou T [ieT^ov eaxw Xey", oxi xal xo A xou Z (ieli^ov eaxai, xav Taov, laov, xav eXaxxov, eXaxxov. Tkel yap ]j.e%6\) eaxi xo A xou T, dXXo 8e xi xo B, xo A dpa Tipoc; xo B jjel^ova Xoyov e^ei f)7iep xo T 7ip6<; xo B. dXX'' [jev xo A Ttpoc; xo B, ouxW(; xo E npoq xo Z, w<; 8e xo T Ttpoc; xo B, avduaXiv ouxwc; xo E icpoc; xo A. xal xo E apa Tipoc; xo Z [lei^ova Xoyov e)(ei f]Tcep xo E Ttpoc; xo A. Ttpoc; o 8e xo auxo jxel^ova Xoyov exei, exelvo eXaaaov eaxiv eXaaaov apa eaxl xo Z xou A- [jel^ov apa eaxl xo A xou Z. 6[jo(cl>c; 8r] 8elc;o[jev, oxi xav Taov fj xo A xCS T, laov eaxai xal xo A xQ Z, xav eXaxxov, eXaxxov. 'Eav apa fj xpia jxeye'dr] xal aXXa auxolc; laa xo TtXfj'doc;, auv8uo Xajjpavojjcva xal ev iu auxw Xoyw, fj 8e xexa- payjjevr] auxwv r\ dvaXoyia, Si'iaou 8e xo TtpQxov xou xpixou jiel^ov fj, xal xo xexapxov xou exxou jiel£ov eaxai, xav taov, nitudes) having a ratio to the same (magnitude), that having the greater ratio is greater [Prop. 5.10]. Thus, D (is) greater than F. Similarly, we can show that even if A is equal to C then D will also be equal to F, and even if (A is) less (than C then D will also be) less (than F). Thus, if there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if), via equality, the first is greater than the third, then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth) . And (if the first is) less (than the third then the fourth will also be) less (than the sixth) . (Which is) the very thing it was required to show. :: e : C then a = 7 as 5 = (. Proposition 21 1 If there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if) their proportion (is) perturbed, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth) . And if (the first is) less (than the third then the fourth will also be) less (than the sixth). A i 1 Di 1 B i 1 E i 1 Ci 1 F i 1 Let A, B, and C be three magnitudes, and D, E, F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two. And let their proportion be perturbed, (so that) as A (is) to B, so E (is) to F, and as B (is) to C, so D (is) to E. And let A be greater than C, via equality. I say that D will also be greater than F. And if (A is) equal (to C then D will also be) equal (to F). And if (A is) less (than C then D will also be) less (than F). For since A is greater than C, and B some other (mag- nitude), A thus has a greater ratio to B than C (has) to B [Prop. 5.8]. But as A (is) to B, so E (is) to F. And, inversely, as C (is) to B, so E (is) to D [Prop. 5.7 corr.]. Thus, E also has a greater ratio to F than E (has) to D [Prop. 5.13]. And that (magnitude) to which the same (magnitude) has a greater ratio is (the) lesser (magni- tude) [Prop. 5.10]. Thus, F is less than D. Thus, D is greater than F. Similarly, we can show that even if A is equal to C then D will also be equal to F, and even if (A is) less (than C then D will also be) less (than F) . 150 ETOIXEIfiN z. ELEMENTS BOOK 5 laov, xdv eXaxxov, eXaxxov onsp eSei SeTi;ai. Thus, if there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if) their proportion (is) per- turbed, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth. And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth). And if (the first is) less (than the third then the fourth will also be) less (than the sixth) . (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : j3 :: e : C and (3 : 7 :: <5 : e then a = 7 as <5 = (. x(3'. 'Edv f) oTtoaaouv jieye'dr] xod aXXa auxolc; !'aa xo TtXfydoc;, auvSuo Xajjpavojjeva xod ev x£> auxw Xoycp, xod 81' taou ev xtp auxw Xoyco eaxai. a ' b ' r>— ' A 1 E ' Z' ' Hi 1 1 K' ' 1 1 Mi 1 1 @l , , A 1 1 ' N' ' ' "Eaxw onoaaoOv ^sys'dr] xd A, B, T xod aXXa auxolc; I'aa xo TtXfj'doc; xa A, E, Z, auvSuo Xa^3otv6^eva ev xG auxcp Xoyw, oiQ \xev xo A npoc; xo B, ouxwc; xo A npbq xo E, cbc; 8e xo B Ttpoc; xo T, ouxwc; xo E Ttpoc; xo Z- Xeyw, oxi xod 81° I'aou ev xw auxw Xoyw eaxai. ElXf](p'f)w yap xwv jxev A, A ig&xu; TtoXXaicXdaia xa H, G, xwv Se B, E aXXa, a exu)(ev, ladxic; icoXXaiiXdaia xa K, A, xod exi xfiv T, Z aXXa, d exu^ev, ladxi? TioXXauXdaia xa M, N. Kal etxel eoxiv d>c; xo A Kpo<; xo B, ouxwg xo A 7ip6<; xo E, xal sl'XrjTCxai xov [lev A, A ladxi? KoXXanXdoia xa H, Q, xwv 8e B, E aXXa, a exu)(ev, ladxic KoXXanXdaia xa K, A, eaxiv dpa d>c xo H TT.poc xo K, ouxw<; xo O npbc, xo A. 8la xd auxd 8r] xal cbc; xo K Kpoc xo M, ouxw<; xo A 7ip6<; xo N. etc! ouv xpia ^teye'dr] eaxl xd H, K, M, xal aXXa auxolc laa xo nkfftoz xa 0, A, N, auv8uo Xa^pavo^teva xal ev tu auxai X6y«, 81' I'aou dpa, ei uuspexet to H xou M, uneps/si xal xo 6 xou N, xa! ei laov, Xaov, xal si eXaxxov, sXaxxov. xal eaxi xd ^.sv H, <d xGv A, A ladxic; noXXanXdoia, xd 8e M, N x£3v r, Z aXXa, a exvyev, ladxic noXXauXdaia. eaxiv dpa &>c, xo A npbc, xo T, oux«<; xo A Tipoc; xo Z. 'Eav dpa f) oiraaaoOv [ieye'dr) xal aXXa auxolc; I'aa xo TiXfj-doc;, auvSuo Xajj.f3av6jj.sva ev ifi auxai Xoyco, xal 8l'Ioou ev x« auxG Xoyco eaxar onep e8ei 8eTc;ai. Proposition 22 f If there are any number of magnitudes whatsoever, and (some) other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality. A 1 B' 1 C ' Di 1 Ei 1 F 1 Gi 1 1 K' 1 1 ' Mi 1 ' Hi 1 1 Li 1 1 1 N 1 1 Let there be any number of magnitudes whatsoever, A, B, C, and (some) other (magnitudes), D, E, F, of equal number to them, (which are) in the same ratio taken two by two, (so that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F. I say that they will also be in the same ratio via equality. (That is, as A is to C, so D is to F.) For let the equal multiples G and H have been taken of A and D (respectively), and the other random equal multiples K and L of B and E (respectively), and the yet other random equal multiples M and N of C and F (respectively) . And since as A is to B, so D (is) to E, and the equal multiples G and H have been taken of A and D (respec- tively), and the other random equal multiples K and L of B and E (respectively), thus as G is to K, so H (is) to L [Prop. 5.4]. And, so, for the same (reasons), as K (is) to M, so L (is) to N. Therefore, since G, K, and M are three magnitudes, and H, L, and other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, thus, via equality, if G exceeds M then H also exceeds N, and if (G is) equal (to M then H is also) equal (to N), and if (G is) less (than M then H is also) less (than N) [Prop. 5.20]. And G and H are equal multiples of A and D (respectively), and M and N other random equal multiples of G and F (respectively) . Thus, as A is to G, so D (is) to F [Def. 5.5]. Thus, if there are any number of magnitudes what- soever, and (some) other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by 151 ETOIXEIfiN z. ELEMENTS BOOK 5 t In modern notation, this proposition reads that if a : /3 :: e : C and /3 : xy'. 'Edv fj xpia ^eyei)/] xal aXXa auxolc; laa to TtXfj'doc; auvSuo Xa^pavo^ieva ev ifi auxco Xoycp, fj 8e xexapayjievr] auxcov fj dvaXoyia, xal 81' laou ev iS auxco Xoycp eaxai. A i 1 b 1 — 1 r ' ' A' — ' Ei 1 Z — ' f-Jl 1 1 1 @ I 1 1 1 J\_l 1 1 K' — i — i — i Mi ' 1 N — i — 1 'Eaxco xp(a ^.eyei!)/] xd A, B, T xal aXXa auxolc; laa xo TtX/j-doc; ouvSuo Xa[ipav6|jieva ev xcp auxcp Xoycp xa A, E, Z, eaxco Be xexapay^ievr] auxcov f) dvaXoyia, (be; ^iev xo A Ttpoc; xo B, ouxcoc; xo E Ttpoc; xo Z, coc; 8s xo B Ttpoc; xo T, ouxcoc; xo A Ttpoc; xo E- Xeyco, oxi eaxlv tlx; xo A Ttpoc; xo T, ouxcoc; xo A Ttpoc; xo Z. EiXfjcpdco xcov [itv A, B, A laaxic; TtoXXaTtXdoia xd H, O, K, xcov Be r, E, Z aXXa, a exu)(ev, lodxic; TtoXXaTtXdoia xd A, M, N. Kal ETtel lodxic; eaxl TtoXXaTtXdaia xd H, O xcov A, B, xd Be \iipr\ xolc; coaauxcoc; TtoXXaTtXaoioic; xov auxov zyz\ Xoyov, eaxiv apa cbc; xo A Ttpoc; xo B, ouxcoc; xo H Ttpoc; xo 6. 8id xa auxd 8f) xal coc; xo E Ttpoc; xo Z, ouxcoc; xo M Ttpoc; xo N - xa[ eaxiv cbc; xo A npoc; xo B, ouxcoc; xo E Ttpoc; xo Z- xal (be; apa xo H Ttpoc; xo 6, ouxcoc; xo M Ttpoc; xo N. xal STtei eaxiv cbc; xo B Ttpoc; xo V, ouxcoc; xo A Ttpoc; xo E, xal evaXXdJ; cbc; xo B Ttpoc; xo A, ouxcoc; xo T Ttpoc; xo E. xal ETtel xd Q, K xebv B, A laaxic; saxl TtoXXaTtXdaia, xd Bs [ispt] xolc; lodxic; TtoXXaTtXaaioic; xov auxov eyei Xoyov, eoxiv apa cbc; xo B Ttpoc; xo A, ouxcoc; xo Ttpoc; xo K. dXX' (be; xo B Ttpoc; xo A, ouxcoc; xo T Ttpoc; xo E- xal cbc; apa xo 6 Ttpoc; xo K, ouxcoc; xo r Ttpoc; xo E. TtdXiv, end xd A, M xcov T, E laaxic; eaxi TtoXXaTtXdaia, eoxiv apa cog xo T Ttpoc; xo E, ouxcoc; xo A Ttpoc; xo M. dXX' (be; xo T Ttpoc; xo E, ouxcoc; xo 9 Ttpoc; xo K' xal cbc; apa xo <d Ttpoc; xo K, ouxcoc; xo A Ttpoc; xo M, xal svaXXdc; cbc; xo <d Ttpoc; xo A, xo K Ttpoc; xo M. eBeix^T) 6e xal cbc; xo H Ttpoc; xo 6, ouxcoc; xo M Ttpoc; xo N. ETtel ouv xpia jieye'dr) taxi xd H, <d, A, xal aXXa auxoic; laa xo TtXfj'dot; xd K, M, N auvBuo Xa[ipav6u.£va ev xco auxco Xoycp, xa[ eoxiv auxcov xexapayu.evr) f) dvaXoyia, 5i' I'oou apa, el UTtepexei xo H xou A, UTtepexei xal xo K xou N, xal ei laov, I'oov, xal ei eXaxxov, eXaxxov. xa[ eaxi xd [lev H, K xcov A, A lodxic; TtoXXaTtXdaia, xd 8e A, N xcov T, Z. eaxiv apa cbc; xo A Ttpoc; xo r, ouxcoc; xo A Ttpoc; xo Z. 'Eav apa fj xpia ^icyc'dr) xal aXXa auxolc; loa xo TtXTjiSoc; auvBuo Xa|ipav6|jieva ev xco auxco Xoycp, fj 8e xexapay^ievr) two, then they will also be in the same ratio via equality. (Which is) the very thing it was required to show. :: C ■ V an d -y ■ S :: rj : 6 then a : 8 :: e : 9. Proposition 23 f If there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality. A 1 C 1 Di — i E' ' F — i G 1 1 1 1 Hi — i — i — 1 Li 1 1 Ki — i — i — i Mi 1 1 Ni — i 1 Let A, B, and C be three magnitudes, and D, E and F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two. And let their proportion be perturbed, (so that) as A (is) to B, so E (is) to F, and as B (is) to C, so D (is) to E. I say that as A is to C, so D (is) to F. Let the equal multiples G, H, and K have been taken of A, B, and D (respectively), and the other random equal multiples L, M, and TV of C, E, and F (respec- tively) . And since G and H are equal multiples of A and B (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as A (is) to B, so G (is) to H. And, so, for the same (reasons), as E (is) to F, so M (is) to N. And as A is to B, so E (is) to F. And, thus, as G (is) to H, so M (is) to N [Prop. 5.11]. And since as B is to C, so D (is) to E, also, alternately, as B (is) to D, so C (is) to E [Prop. 5.16]. And since H and K are equal multiples of B and D (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as B is to D, so H (is) to K. But, as B (is) to D, so C (is) to E. And, thus, as H (is) to K, so C (is) to E [Prop. 5.11]. Again, since L and M are equal multiples of C and E (re- spectively), thus as C is to E, so L (is) to M [Prop. 5.15]. But, as C (is) to E, so H (is) to K. And, thus, as H (is) to K, so L (is) to M [Prop. 5.11]. Also, alternately, as H (is) to L, so K (is) to M [Prop. 5.16]. And it was also shown (that) as G (is) to H, so M (is) to N. Therefore, since G, H, and L are three magnitudes, and K, M, and N other (magnitudes) of equal number to them, (being) in the same ratio taken two by two, and their proportion is perturbed, thus, via equality, if G exceeds L then K also exceeds N, and if (G is) equal (to L then K is also) equal (to N), and if (G is) less (than L then K is also) less (than N) [Prop. 5.21]. And G and K are equal mul- tiples of A and D (respectively), and L and N of G and 152 ETOIXEIfiN z. ELEMENTS BOOK 5 auxcov f] dvaXoyia, xal 8i'foou ev xcp aOxcp Xoycp eaxai- onep F (respectively). Thus, as ^4 (is) to C, so D (is) to F eSei 8el£ai. [Def. 5.5]. Thus, if there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : /3 :: e : £ and (3 : 7 :: 8 : t then a : 7 :: <5 : ( . x8'. 'Edv uptOTOv npbz Seuxepov xov auxov e)(r) Xoyov xod xpixov 7ip6<; xexapxov, exn 8s xal TtefiTixov npbz Seuxepov xov auxov Xoyov xal exxov 7ip6<; xexapxov, xal auvxeiDev TtpGxov xal TtefjiTCXov Tipoc; Seuxepov xov auxov ec;ei Xoyov xal xpixov xal exxov 7ip6<; xexapxov. B A' ' 'H r> 1 E A 1 1 © Z' 1 IlpGxov yap xo AB Txpoc; Seupepov xo T xov auxov exexw Xoyov xal xpixov xo AE Tipoc; xexapxov xo Z, exexco Se xal Tte^iTtxov xo BH Tipoc; Seuxepov xo T xov auxov Xoyov xal exxov xo E0 Tipoc; xexapxov xo Z- Xeyw, oxi xal auvxcdev TipaSxov xal Tiefjmxov xo AH Tipoc; Seuxepov xo T xov auxov e^ei Xoyov, xal xpixov xal exxov xo A0 Tipoc; xexapxov xo Z. Tkel yap eaxiv cdc, xo BH Tipoc; xo T, ouxcoc; xo E0 Tipoc; xo Z, dvaTiaXiv apa oX xo T Tipoc; xo BH, ouxgk xo Z Tipoc; xo E9. etc el ouv eaxiv oX xo AB Tipoc; xo T, ouxgjc; xo AE Tipoc; xo Z, oX Se xo T Tipoc; xo BH, ouxck xo Z Tipoc; xo E6, Si' 'laou dpa eaxiv oX xo AB Tipoc; xo BH, ouxgjc; xo AE Tipoc; xo E0. xal ctccI Sir]pr)fisva ^teyei!)/] dvdXoyov eaxiv, xal auvxcdevxa dvdXoyov eaxai- eaxiv apa oX xo AH Tipoc; xo HB, ouxck xo A9 npbz xo 9E. eaxi Se xal 6X xo BH Tipoc; xo T, ouxcx xo E6 Tipoc; xo Z- 81' laou apa eaxiv cX xo AH Tipoc; xo T, ouxgjc; xo A9 Tipoc; xo Z. 'Edv dpa upwxov Tipoc; Seuxepov xov auxov e^T) Xoyov xal xpixov Tipoc; xexapxov, e^T] Se xal tic^ltixov Tipoc; Seuxepov xov auxov Xoyov xal exxov Tipoc; xexapxov, xal auvxeiDev TtpaSxov xal Tiefjmxov Tipoc Seuxepov xov auxov e^ei Xoyov xal xpixov xal exxov Tipoc; xexapxov oTiep eSei Selc;ai. Proposition 24 f If a first (magnitude) has to a second the same ratio that third (has) to a fourth, and a fifth (magnitude) also has to the second the same ratio that a sixth (has) to the fourth, then the first (magnitude) and the fifth, added together, will also have the same ratio to the second that the third (magnitude) and sixth (added together, have) to the fourth. B A 1 1 G Ci 1 E Di 1 H F 1 1 For let a first (magnitude) AB have the same ratio to a second C that a third DE (has) to a fourth F. And let a fifth (magnitude) BG also have the same ratio to the second C that a sixth EH (has) to the fourth F. I say that the first (magnitude) and the fifth, added together, AG, will also have the same ratio to the second C that the third (magnitude) and the sixth, (added together), DH, (has) to the fourth F. For since as BG is to C, so EH (is) to F, thus, in- versely, as C (is) to BG, so F (is) to EH [Prop. 5.7 corr.]. Therefore, since as AB is to C, so DE (is) to F, and as C (is) to BG, so F (is) to EH, thus, via equality, as AB is to BG, so DE (is) to EH [Prop. 5.22]. And since separated magnitudes are proportional then they will also be pro- portional (when) composed [Prop. 5.18]. Thus, as AG is to GB, so DH (is) to HE. And, also, as BG is to C, so EH (is) to F. Thus, via equality as AG is to C, so DH (is) toF [Prop. 5.22]. Thus, if a first (magnitude) has to a second the same ratio that a third (has) to a fourth, and a fifth (magni- tude) also has to the second the same ratio that a sixth (has) to the fourth, then the first (magnitude) and the fifth, added together, will also have the same ratio to the second that the third (magnitude) and the sixth (added 153 ETOIXEIfiN z. ELEMENTS BOOK 5 together, have) to the fourth. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : (3 :: 7 : 8 and e : f3 :: ( : 5 then a + e : f3 :: 7 + £ : 5. XZ 'Edv Tsaaapa ^ley^TI dvdXoyov rj, to ^eyiaTov [auxcov] xal to eXd/iaTov Suo tcov Xoituov ^ei^ovd eoxiv. Ah Eh H — 1 — Zh — I — h A T5ax« xeaaapa (j.eye'd'ir) dvdXoyov id AB, IA, E, Z, a>c to AB Ttpoc; to TA, outw<; to E itpoi; to Z, eoto Se ^isyioTov [Lev ocutwv to AB, eXd)(icn:ov 8e to Z- Xeyw, oti Ta AB, Z twv TA, E [isi^ovd eaTiv. Kslcrdco yap T 9 E laov to AH, ifi Se Z laov to TO. 'End [ouv] eaTiv £>c to AB 7tp6<; to TA, outck to E Ttpoc; to Z, laov 8s to [iev E t£) AH, to 8e Z tG r9, eaTiv dpa foe. to AB npoc; to TA, outgjc; to AH 7ipo<; to T0. xal ensi eaTiv foe, oXov to AB 7ip6<; 6Xov to TA, outca; dcpaipcdev to AH npoc; dcpaipcdev to I?0, xal Xoitcov dpa to HB npbe, Xoikov to OA ecrcai foe, oXov to AB Ttpoc; oXov to TA. ^ta^ov 8e to AB tou TA- uelCov dpa xal to HB toO 9A. xal enel laov eaTi to [iev AH tw E, to Se T0 t« Z, Ta dpa AH, Z laa IcttI toI<; T9, E. xal [enel] eav [dviaou; I'aa TipoaT£i!)fi, Ta oXa dviad eaTiv, eav dpa] twv HB, 0A dviawv ovtwv xal jieiC^ovoc; toO HB tw ^tev HB TtpoaTcdfj Ta AH, Z, tG Se 9A TipooTei9f) Ta TO, E, auvdyeTai Ta AB, Z jieiCova twv TA, E. 'Eav dpa Teaaapa jieye'dr) dvdXoyov fj, to jieyiaTOv auTSSv xal to eXd)(iaTov 860 tc5v XoituSv ^eii^ovd eaTiv. oitep eSei 8eT^ai. Proposition 25t If four magnitudes are proportional then the (sum of the) largest and the smallest [of them] is greater than the (sum of the) remaining two (magnitudes) . G 1 Ah Eh o H —1— t 1 1 Let AB, CD, E, and F be four proportional magni- tudes, (such that) as AB (is) to CD, so E (is) to F. And let AB be the greatest of them, and F the least. I say that AB and F is greater than CD and E. For let AG be made equal to E, and CH equal to F. [In fact,] since as AB is to CD, so i? (is) to F, and £ (is) equal to AG, and F to CH, thus as is to CD, so AG (is) to CH. And since the whole AB is to the whole CD as the (part) taken away AG (is) to the (part) taken away CH, thus the remainder GB will also be to the remainder HD as the whole AB (is) to the whole CD [Prop. 5.19]. And AB (is) greater than CD. Thus, GB (is) also greater than HD. And since AG is equal to E, and GH to F, thus AG and F is equal to G# and E. And [since] if [equal (magnitudes) are added to unequal (magnitudes) then the wholes are unequal, thus if] AG and F are added to GB, and CH and £ to HD—GB and iJZ? being unequal, and GB greater — it is inferred that AB and F (is) greater than CD and E. Thus, if four magnitudes are proportional then the (sum of the) largest and the smallest of them is greater than the (sum of the) remaining two (magnitudes). (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : (3 :: 7 : S, and a is the greatest and <5 the least, then a + 5 > /3 + 7. 154 ELEMENTS BOOK 6 Similar Figures 155 ETOIXEIfiN 9'. ELEMENTS BOOK 6 "Opoi. a'. "Ojioia ax^axa £U , duypa|j.[id eaxiv, oaa xdc; xe ycoviac; laac; e/ei xaxa ^nav xal xdc; Ticpl xdc; Taac; ycoviac; TtXeupdc; dvdXoyov. P'. 'Axpov xal [isaov Xoyov eu-deia xsx^ifjadai Xeyexai, oxav rj to? f) oXr) Ttpoc; xo ^si^ov x^trjua, ouxtoc; xo ^isTCov Ttpoc; xo eXaxxov. y'. "Ytyoz eaxi ndvxoc; axr^axoc; r) and xrjc; xopucprjc; era xr)v pdaiv xd-dexoc; dyo[ievr). a . Td xpiycova xdi xd 7tapaXX/]X6ypa|i^a xd bub xo auxo ix\)oc, ovxa izpbc, dXX/jXd eoxiv tbc; di pdaeic;. E A Z © H B T A K A "Eot(x> xpiycova piev xd ABr, ArA, TtapaXXr)X6ypa[i(jia 8s xd Er, rZ bub xo auxo uijjoc; xo AE Xeyco, oxi eoxlv coc; f] Br pdoig Ttpoc; xt)v TA pdaic;, ouxcoc; xo ABr xpiycovov Ttpoc; xo ArA xpiycovov, xal xo Er 7iapaXXr)X6ypa^ov npbc, xo rZ TtapaXXr]X6ypa^ov. 'ExpepXr]a"d« yap i] BA ecp' exdxepa xd [ispt] era xd 0, A a/jjiaa, xal xricrdcoaav xrj ^iev Br pdaei Taai [6oai8r)noxo0v] ai BH, H6, xrj 8s TA pdaei laai oaaiBrjTioxouv ai AK, KA, xal eTTe^euyjdcoaav ai AH, A6, AK, AA. Kal ETZsi laai eialv ai TB, BH, H9 dXXr|Xaic;, laa eaxi xal xd AOH, AHB, ABr xpiycova dXXr|Xoic;. 6aa7tXaaicov apa eaxlv f) OF pdaig xrjc; Br pdaecoc;, xoaauxanXdoiov eaxi xal xo A0r xpiycovov xou ABr xpiycovou. 8id xd auxa 8/) oaarcXaaicov eaxlv f] Ar pdaic; xrjc; TA pdaecoc;, xoaau- xanXdaiov eaxi xal xo AAr xpiycovov xou ArA xpiycovou- xal et Xar] eaxlv f] 6r pdaic; xrj IA pdaei, laov eaxi xal xo AOr xpiycovov xtp ArA xpiytovto, xal si U7iepe)(ei f] ®T pdaic; xrjc; TA pdaecoc;, UTtepe/ei xal xo A0r xpiycovov xou ArA xpiycovou, xal ei eXdaatov, eXaaaov. xeaadptov 8r) ovxtov ^.eye'dcov 8uo [iev pdaecov xcov Br, TA, Suo Se xpiycovcov xcov ABr, ArA elXrjuxai iadxic; uoXXaTiXdaia xrjc; (iev Br pdaecoc; xal xou ABr xpiycovou fj xe 0r pdaic; xal xo A6r xpiycovov, xrjc; 8s TA pdaecoc; xal xou AAr xpiycovou dXXa, Definitions 1. Similar rectilinear figures are those (which) have (their) angles separately equal and the (corresponding) sides about the equal angles proportional. 2. A straight-line is said to have been cut in extreme and mean ratio when as the whole is to the greater seg- ment so the greater (segment is) to the lesser. 3. The height of any figure is the (straight-line) drawn from the vertex perpendicular to the base. Proposition V Triangles and parallelograms which are of the same height are to one another as their bases. E A F H G B C D K L Let ABC and ACD be triangles, and EC and CF par- allelograms, of the same height AC. I say that as base BC is to base CD, so triangle ABC (is) to triangle ACD, and parallelogram EC to parallelogram CF. For let the (straight-line) BD have been produced in each direction to points H and L, and let [any number] (of straight-lines) BG and GH be made equal to base BC, and any number (of straight-lines) DK and KL equal to base CD. And let AG, AH, AK, and AL have been joined. And since CB, BG, and GH are equal to one another, triangles AHG, AGB, and ABC are also equal to one another [Prop. 1.38]. Thus, as many times as base HC is (divisible by) base BC, so many times is triangle AHC also (divisible) by triangle ABC. So, for the same (rea- sons), as many times as base LC is (divisible) by base CD, so many times is triangle ALC also (divisible) by triangle ACD. And if base HC is equal to base CL then triangle AHC is also equal to triangle ACL [Prop. 1.38]. And if base HC exceeds base CL then triangle AHC also exceeds triangle ACL} And if (HC is) less (than CL then AHC is also) less (than ACL). So, their being four magnitudes, two bases, BC and CD, and two trian- 156 ETOIXEIfiN 9'. ELEMENTS BOOK 6 a exuxev, ladxic TtoXXarcXaaia fj xe Ar pdcnc xal to AAr xplyovov xal 8e8eixxai, oxi, ei uitepexei T) Or pdaic; xfj? IA pdaeoK, UTtepexei xal xo A0I 1 xpiycovov xou AAr xpiycivou, xai ei lor], I'aov, xdi ei eXaaawv, eXaaaov eaxiv dpa &>z rj Br pdai? upoc; xrjv EA pdaiv, oux«<; xo ABr xplyovov 7tpo<; xo ArA xplywvov. Kdi ctieI xou [ibv ABr xpiywvou BrnXdaiov eaxi xo Er 7tapaXX/]X6ypa^ov, xou 5e ArA xpiyovou BinXdaiov eaxi xo Zr 7tapaXX/]X6ypa^ov, xd 8e jiepr) xou; cbaauxco? 710X- XaTtXaaioic; xov auxov exei Xoyov, eaxiv dpa aX xo ABr xplyovov Kp6<; xo ArA xpiycovov, ouxoc xo Er TtapaX- XrjXoypa^ov Ttpoc; xo ZT TtapaXXrjXoypa^ov. enel ouv eSeix^T), ok (j.ev rj Br pdai<; n;p6<; x/]v EA, ouxgx xo ABr xplyovov upoc; xo ArA xplywvov, cbc; Se xo ABr xpiycovov 7ip6<; xo ArA xpiywvov, ouxco? xo Er 7iapaXXr]X6ypa^ov Tipoc xo rZ TtapaXXrjXoypa^iov, xal &<; dpa f] Br pdau; 7ip6<; xr]v TA pdaiv, ouxwc; xo Er 7tapaXX/]X6ypa[i^ov Ttpoc; xo Zr TtapaXXrjXoypa^ov. Td dpa xpiycova xal xd TtapaXXr)X6ypa^a xd O716 xo auxo utjioc; ovxa npoc dXXrjXd eaxiv cbt; al pdaeic oTtep e8ei 8eTc;ai. gles, ABC and ACD, equal multiples have been taken of base BC and triangle ABC — (namely), base HC and tri- angle AH C — and other random equal multiples of base CD and triangle ADC — (namely), base LC and triangle ALC. And it has been shown that if base HC exceeds base CL then triangle AHC also exceeds triangle ALC, and if {HC is) equal (to CL then AHC is also) equal (to ALC), and if {HC is) less (than CL then AHC is also) less (than ALC). Thus, as base BC is to base CD, so triangle ABC (is) to triangle ACD [Def. 5.5]. And since parallelogram EC is double triangle ABC, and parallelo- gram FC is double triangle ACD [Prop. 1.34], and parts have the same ratio as similar multiples [Prop. 5.15], thus as triangle ABC is to triangle ACD, so parallelogram EC (is) to parallelogram FC. In fact, since it was shown that as base BC (is) to CD, so triangle ABC (is) to triangle ACD, and as triangle ABC (is) to triangle ACD, so par- allelogram EC (is) to parallelogram CF, thus, also, as base BC (is) to base CD, so parallelogram EC (is) to parallelogram FC [Prop. 5.11]. Thus, triangles and parallelograms which are of the same height are to one another as their bases. (Which is) the very thing it was required to show. t As is easily demonstrated, this proposition holds even when the triangles, or parallelograms, do not share a common side, and/or are not right-angled. t This is a straight-forward generalization of Prop. 1.38. P'- 'Edv xpiyovou Ttapd ^uav xwv TtXeupov dx^rj xu; eu'deTa, dvdXoyov xejiel xd<; xou xpiywvou TtXeupdc;- xal edv ai xou xpiyovou TiXeupal dvdXoyov xu/jiSwcnv, i\ iia xcic, xo^iac; era- ^euyvu^ievr] eu-dela Ttapd xr)v Xoin/jv eaxai xou xpiycovou TiXeupdv. A Tpiyiovou yap xou ABr 7xapdXXr)Xo<; [iia xcov nXeupfiv xfj Br fix'dw f) AE- Xeyw, oxi eaxiv «<; f] BA Ttp6<; x/]v AA, ouxdK f) TE Ttp6<; xrjv EA. Proposition 2 If some straight-line is drawn parallel to one of the sides of a triangle then it will cut the (other) sides of the triangle proportionally. And if (two of) the sides of a tri- angle are cut proportionally then the straight-line joining the cutting (points) will be parallel to the remaining side of the triangle. A For let DE have been drawn parallel to one of the sides BC of triangle ABC. I say that as BD is to DA, so CE (is) to EA. 157 ETOIXEIfiN 9'. ELEMENTS BOOK 6 'ETte^eu/'dcoaav yap ai BE, IA. laov apa eaxl to BAE xpiycovov iS IAE xpiycbvy era yap xrjc auxfjc; pdae(bc; eaxi xfjc; AE xal ev xalc auxaTc TtapaXXfjXoic; xdic; AE, BE dXXo 8e ti to AAE xpiywvov. xd 8s Taa Ttpoc; to auxo xov auxov e)(ei Xoyov eaxiv apa (be; xo BAE xpiywvov Tipoc; xo AAE [xpiyovov], ouxwc; xo IAE xpiywvov Ttpoc; xo AAE xpiyovov. aXX' (be; [lev xo BAE xpiycovov Ttpoc; xo AAE, ouxwc; fj BA Tipoc; xrjv AA- utio yap xo auxo utjioc; ovxa xrjv aTto xou E stcl xrjv AB xd-dexov dyo^ievrjv Tipoc; dXXrjXd eiaiv (be; ai pdaeic- 5ia xa auxa 8rj (be; xo TAE xpiyovov Tipoc; xo AAE, ouxcx; fj TE Ttpoc; xrjv EA- xal (be; apa rj BA Tipoc; xf]v AA, ouxog fj TE Tipoc; xrjv EA. AXXd 8rj ai xou ABr xpiytbvou TtXeupal ai AB, Ar dvdXoyov xexurjadtoaav, (be; fj BA Tipoc; xrjv AA, ouxcoc f] TE Ttpoc; xrjv EA, xal CTte£e6)(i!)Gj f) AE- Xeyto, oxi TiapdXXrjXoc; eaxiv fj AE xfj Br. Tcbv yap auxfiv xaxaaxeuaa-devxwv, enei eaxiv (be; fj BA Tipoc; xrjv AA, ouxgjc; fj TE Tipoc; xrjv EA, dXX' (be; (lev f] BA Ttpoc; xrjv AA, ouxwc; xo BAE xpiywvov Ttpoc xo AAE xpiycovov, (be; Be rj TE Tipoc; xrjv EA, ouxwc; xo TAE xpiywvov Ttpoc; xo AAE xpiywvov, xal (be; apa xo BAE xpiywvov Ttpoc; xo AAE xpiywvov, ouxwc; xo TAE xpiyovov Ttpoc; xo AAE xpiywvov. exdxepov apa xwv BAE, TAE xpiywvwv Ttpoc; xo AAE xov auxov e)(ei Xoyov. taov apa eaxl xo BAE xpiywvov iu TAE xpiywvy xai eiaiv excl xrjc; auxfjc; pdae«<; xfjc; AE. xa 8e laa xpiywva xal era xfjc; auxfjc pdaewc ovxa xal ev xdic auxdic TtapaXXfjXoic eaxiv. TtapdXXrjXoc apa eaxlv fj AE xrj Br. 'Eav apa xpiywvou itapa (liav xwv TtXeupwv dx'dfj xic euif)eTa, dvdXoyov xe^tel xac xou xpiywvou TtXeupdc xal eav ai xou xpiywvou itXeupai dvdXoyov xjirj'dGaiv, fj era xac; xoiudc eraCeuyvu^ievrj euifteTa Ttapa xrjv XoiTtfjv eaxai xou xpiyovou TtXeupdv oTtep e8ei 8e1c;ai. Y • 'Eav xpiywvou fj ywvia 8[)(a xurj'dfj, fj 8e xe[Uvouaa xrjv ywviav eO'dela xejivrj xal xrjv pdaiv, xa xfjc; pdaewc; x^irj^axa xov auxov ec;ei Xoyov xdic; XoiTtdic; xou xpiywvou TtXeupaTc;- xal eav xd xfjc; pdoewc; x^irjjiaxa xov auxov g/rj Xoyov xalc; Xomalc; xou xpiycovou TtXeupaTc;, fj dito xfjc; xopucpfjc; era xrjv xojurjv era^euyvu[uevrj eu'dela 8i/a xejiel xrjv xou xpiycovou ycoviav. "Eot(x> xpiywvov xo ABr, xal xexjurjo'dw fj UTto BAr ywvia 8i)(a UTto xrjc; AA eu-deiac;- Xey«, oxi eaxlv (be; fj BA Ttpoc; xrjv TA, ouxoc; fj BA Ttpoc; xrjv Ar. "H)cd« yap Bid xou T xrj AA TtapdXXrjXoc; rj TE, xal 8ia)cdeTaa fj BA aujiTtiTtxexcd auxfj xaxa xo E. For let BE and CD have been joined. Thus, triangle BDE is equal to triangle CDE. For they are on the same base DE and between the same parallels DE and BC [Prop. 1.38]. And ADE is some other triangle. And equal (magnitudes) have the same ra- tio to the same (magnitude) [Prop. 5.7]. Thus, as triangle BDE is to [triangle] ADE, so triangle CDE (is) to trian- gle ADE. But, as triangle BDE (is) to triangle ADE, so (is) BD to DA. For, having the same height — (namely), the (straight-line) drawn from E perpendicular to AB — they are to one another as their bases [Prop. 6.1]. So, for the same (reasons), as triangle CDE (is) to ADE, so CE (is) to EA. And, thus, as BD (is) to DA, so CE (is) to EA [Prop. 5.11]. And so, let the sides AB and AC of triangle ABC have been cut proportionally (such that) as BD (is) to DA, so CE (is) to EA. And let DE have been joined. I say that DE is parallel to BC. For, by the same construction, since as BD is to DA, so CE (is) to EA, but as BD (is) to DA, so triangle BDE (is) to triangle ADE, and as CE (is) to EA, so triangle CDE (is) to triangle ADE [Prop. 6.1], thus, also, as tri- angle BDE (is) to triangle ADE, so triangle CDE (is) to triangle ADE [Prop. 5.11]. Thus, triangles BDE and CDE each have the same ratio to ADE. Thus, triangle BDE is equal to triangle CDE [Prop. 5.9]. And they are on the same base DE. And equal triangles, which are also on the same base, are also between the same paral- lels [Prop. 1.39]. Thus, DE is parallel to BC. Thus, if some straight-line is drawn parallel to one of the sides of a triangle, then it will cut the (other) sides of the triangle proportionally. And if (two of) the sides of a triangle are cut proportionally, then the straight-line joining the cutting (points) will be parallel to the remain- ing side of the triangle. (Which is) the very thing it was required to show. Proposition 3 If an angle of a triangle is cut in half, and the straight- line cutting the angle also cuts the base, then the seg- ments of the base will have the same ratio as the remain- ing sides of the triangle. And if the segments of the base have the same ratio as the remaining sides of the trian- gle, then the straight-line joining the vertex to the cutting (point) will cut the angle of the triangle in half. Let ABC be a triangle. And let the angle BAC have been cut in half by the straight-line AD. I say that as BD is to CD, so BA (is) to AC. For let CE have been drawn through (point) C par- allel to DA. And, BA being drawn through, let it meet 158 ETOIXEIfiN 9'. ELEMENTS BOOK 6 E B A r Kal excel sic; rcapaXXf]Xouc; xdc; AA, Er eudela everceaev f) Ar, f] apa utio ArE ytovia Tar) eaxl xfj Oreo TAA. dXX' f] utio TAA xfj Oreo BAA urcoxeixai Tar)- xal f] utio BAA apa xfj utio ArE eaxiv Tar). rcdXiv, creel eic, TiapaXXfjXouc; xdc; AA, Er £1)15610 eveiieaev f] BAE, f) exxoc; ywvia f] Oreo BAA Tar) eaxl xfj evxoc; xfj utco AEr. sSei/i}/) 8s xal f] utio ArE xfj utco BAA Tar)- xal f) utco ArE dpa ycovia xfj utco AEr eaxiv Tar)- &axe xal TcXeupd f| AE TiXeupa xfj Ar eaxiv Xor\. xal eicel xpiycovou xou BrE Tcapd ^iav xwv TcXeupwv xf)v Er fjxxai f) AA, dvdXoyov apa eaxiv (be; f] BA Tcpoc; xf)v Ar, ouxwc; f) BA icpoc; xf)v AE. Tar] Be f) AE xfj Ar- cbc; apa f] BA Ttpoc; xf)v Ar, oux«<; f] BA icpoc; xf)v Ar. AXXd 8fj eaxco 6c; f] BA Tcpoc; xf]v Ar, ouxwc; f] BA icpoc xf]v Ar, xal iTieCeux'dw f] AA- Xeyto, oxi Biya xex^rjxai f| utco BAr ycovia utco xfjc; AA eufteiac;. Tt5v yap auxwv xaxaaxeuaa'devxwv, CTcei eaxiv «<; f) BA Tcpoc; xf]v Ar, ouxwc; f) BA Tcpoc; xf]v Ar, dXXd xal wc; f] BA Tcpoc; xr)v Ar, ouxwc; eaxiv f] BA Tcpoc; xf]v AE- xpiycovou yap xou BrE Tcapd ulav xf]v Er fjxxai f] AA- xal dbc; apa f] BA Tcpoc; xf]v Ar, ouxwc; f] BA Tcpoc; xf)v AE. Tar) apa f] Ar xfj AE- 6axe xal ywvia f) utco AEr xfj utco ArE eaxiv Tar). dXX' f] ^xev utco AEr xfj exxoc; xfj utio BAA [eaxiv] Tar], f) 8e utco ArE xfj evaXXdc; xfj utco TAA eaxiv Tar)- xal f) utio BAA apa xfj utco TAA eaxiv Tar), f) apa utco BAr yovia 8i)(a xexurjxai utio xfjc; AA eu-Mac;. 'Eav dpa xpiycivou f] ywvia 8i)(a x^trydfj, f) 5e xe^ivouaa xf)V ywviav eu'dela xe^ivr] xal xf]v pdaiv, xa xfjc; pdaewc; x^ifjuaxa xov auxov ei;ei Xoyov xalc; Xomau; xou xpiywvou TiXeupaTc;- xal edv xa xfjc; pdaeioc; xjifj^iaxa xov auxov ey^r) Xoyov xalc XoiTtau; xou xpiywvou nXeupdic;, f) dTco xfjc; xo- pucpfjc; era xf)v xo^tfjv eTci£euyvutJievr] eu-dela 5ixa xeuvei xf)v xou xpiycovou ywviav onep eSei 8eTc;ai. t The fact that the two straight-lines meet follows because the sum of (CE) at (point) El E B DC And since the straight-line AC falls across the parallel (straight-lines) AD and EC, angle ACE is thus equal to CAD [Prop. 1.29]. But, (angle) CAD is assumed (to be) equal to BAD. Thus, (angle) BAD is also equal to ACE. Again, since the straight-line BAE falls across the parallel (straight-lines) AD and EC, the external angle BAD is equal to the internal (angle) AEC [Prop. 1.29]. And (angle) ACE was also shown (to be) equal to BAD. Thus, angle ACE is also equal to AEC. And, hence, side AE is equal to side AC [Prop. 1.6]. And since AD has been drawn parallel to one of the sides EC of triangle BCE, thus, proportionally, as BD is to DC, so BA (is) to AE [Prop. 6.2]. And AE (is) equal to AC. Thus, as BD (is) to DC, so BA (is) to AC. And so, let BD be to DC, as BA (is) to AC. And let AD have been joined. I say that angle BAG has been cut in half by the straight-line AD. For, by the same construction, since as BD is to DC, so BA (is) to AC, then also as BD (is) to DC, so BA is to A_E. For AD has been drawn parallel to one (of the sides) EC of triangle BCE [Prop. 6.2]. Thus, also, as BA (is) to AC, so BA (is) to AE [Prop. 5.11]. Thus, AC (is) equal to AE [Prop. 5.9]. And, hence, angle AEC is equal to ACE [Prop. 1.5]. But, AEC [is] equal to the external (angle) BAD, and ACE is equal to the alternate (angle) CAD [Prop. 1.29]. Thus, (angle) BAD is also equal to CAD. Thus, angle BAG has been cut in half by the straight-line AD. Thus, if an angle of a triangle is cut in half, and the straight-line cutting the angle also cuts the base, then the segments of the base will have the same ratio as the re- maining sides of the triangle. And if the segments of the base have the same ratio as the remaining sides of the triangle, then the straight-line joining the vertex to the cutting (point) will cut the angle of the triangle in half. (Which is) the very thing it was required to show. E and CAE is less than two right-angles, as can easily be demonstrated. 159 ETOIXEIfiN 9'. ELEMENTS BOOK 6 See Post. 5. 5'. Ttov iaoycovicov xpiytovtov dvdXoyov eiaiv ai TtXeupai al Tiepi xdc; loac ytoviac; xai o^oXoyoi ai utio xdc; iaac; ycoviac; UTioxdvouaai. z b r e "Ecrao iaoya>via xpiywva xd ABr, ArE iarjv £)(ovxa xrjv ^iev utio ABr ycoviav xrj utio ArE, x/)v 8e utio BAr xfj utio TAE xai exi xrjv utio ArB xfj utio TEA- Xeyto, oxi xiov ABr, ArE xpiytoviov dvdXoyov eiaiv ai TiXsupai ai Tiepi xdc iaac; ycoviac; xai 6[i6Xoyoi ai (mo xdc; iaac; ycoviac; UTCoxdvouaai. Ksia'dco yap in euiiteiac; f) Br xfj TE. xai enel ai utio ABr, ArB ycoviai 8uo 6pi5cov sXdxxovec; eiaiv, iar] 8e f] utio ArB xrj utio AEr, ai apa utio ABr, AEr 8uo opiScov eXdxxovec; sioiv ai BA, EA apa £xpaXX6|ievai au^i- Tteaouvxai. expepXf|a , dcoaav xai au^iTiiTixexcoaav xaxd xo Z. Kai stcei lot) eaxiv f] utio ArE ycovia xrj utio ABr, TiapdXXrjXoc; eaxiv r) BZ xfj TA. TidXiv, ETcei iar) Eaxiv f] utio ArB xfj utio AEr, TiapdXXrjXoc; eaxiv f] AT xrj ZE. TiapaX- Xr]X6ypa[i|jiov apa eaxi xo ZAIA- iar) apa f) uev ZA xfj Ar, f] 8s Ar xrj ZA. xai ercei xpiycovou xou ZBE napd (iiav xrjv ZE rjxxai f) Ar, eaxiv apa cbc; f] BA Tipoc; xr)v AZ, ouxcoc; i\ Br Tipoc; xr]v TE. iar) 8e f) AZ xfj IA- 6c; apa f) BA Tipoc; xr)v IA, ouxcoc; f) Br Tipoc; xr)v TE, xai evaXXdi; cbc; f) AB Tipoc xr)v Br, ouxcoc; fj Ar Tipoc; xrjv TE. TidXiv, enei TiapdXXrjXoc; Eaxiv f) TA xfj BZ, Eaxiv apa cbc; f) Br Tipoc; xrjv TE, ouxcoc; f) ZA Tipoc; xrjv AE. i'arj 8s f) ZA xfj Ar- cbc; apa fj Br Tipoc; xrjv TE, ouxcoc; fj Ar Tipoc; xrjv AE, xai svaXXac; cbc; f) Br Tipoc; xrjv TA, ouxcoc; fj TE Tipoc; xrjv EA. eTiei oOv e8ei)($r) cbc [i£v f) AB Tipoc; xrjv Br, ouxcoc; fj Ar Tipoc; xrjv TE, cbc; 8s rj Br Tipoc; xrjv TA, ouxcoc; fj TE Tipoc; xrjv EA, 8i' iaou apa cog f] BA Tipoc; xrjv Ar, ouxcoc; f] TA Tipoc; xrjv AE. Tcov apa iaoycovicov xpiycovcov dvdXoyov eiaiv ai TtXeupai ai Tiepi xdc; iaac; ycoviac; xai o^toXoyoi ai utio xdc; iaac; ycoviac; UTioxsivouaai- oTiep e8ei SeTc;ai. Proposition 4 In equiangular triangles the sides about the equal an- gles are proportional, and those (sides) subtending equal angles correspond. F B C E Let ABC and DCE be equiangular triangles, having angle ABC equal to DCE, and (angle) BAG to CDE, and, further, (angle) ACB to CED. I say that in trian- gles ABC and DCE the sides about the equal angles are proportional, and those (sides) subtending equal angles correspond. Let BC be placed straight-on to CE. And since angles ABC and ACB are less than two right-angles [Prop 1.17], and ACB (is) equal to DEC, thus ABC and DEC are less than two right-angles. Thus, BA and ED, being produced, will meet [C.N. 5]. Let them have been produced, and let them meet at (point) F. And since angle DCE is equal to ABC, BF is parallel to CD [Prop. 1.28]. Again, since (angle) ACB is equal to DEC, AC is parallel to EE [Prop. 1.28]. Thus, FACD is a parallelogram. Thus, FA is equal to DC, and AC to FD [Prop. 1.34]. And since AC has been drawn parallel to one (of the sides) FE of triangle FBE, thus as BA is to AF, so BC (is) to CE [Prop. 6.2]. And AF (is) equal to CD. Thus, as BA (is) to CD, so BC (is) to CE, and, alternately, as AB (is) to BC, so DC (is) to CE [Prop. 5.16]. Again, since CD is parallel to BF, thus as BC (is) to CE, so FD (is) to DE [Prop. 6.2]. And FD (is) equal to AC. Thus, as BC is to CE, so AC (is) to DE, and, alternately, as BC (is) to CA, so Ci? (is) to ED [Prop. 6.2]. Therefore, since it was shown that as AB (is) to BC, so DC (is) to CE, and as BC (is) to CA, so Ci? (is) to ED, thus, via equality, as BA (is) to AC, so CD (is) to [Prop. 5.22]. Thus, in equiangular triangles the sides about the equal angles are proportional, and those (sides) subtend- 160 ETOIXEIfiN 9'. ELEMENTS BOOK 6 z. 'Edv 80o xptyova xdc; TiXeupdc; dvdXoyov e)(r), iaoyiovia saxai xd xpiycova xal foac; e^si xdc; yojviac;, uc p' ac; at 6(i6Xoyoi TiXeupal UKOxeivouaiv. A A "Eaxco 860 xpiyova xd ABr, AEZ xdc; TiXsupdc; dvdXoyov e^ovxa, (be; ^lev xrjv AB Tipoc; xf)v Br, ouxwc; xf)v AE Tipoc; xf)v EZ, (be Ss xrjv Br Tipoc; xf)v FA, ouxwc; xf)v EZ Tipoc; xf)v ZA, xal ixi (be; xrjv BA Tipoc; xf]v Ar, ouxck xf]v EA Tipoc; xrjv AZ. Xeyw, oxi fooyoviov eaxi xo ABr xpiywvov iw AEZ xpiycovw xal foac; e^ouai xdc; ywviac;, Ocp' ac; ai o^ioXoyoi TiXeupal UTioxeivouaiv, xrjv ^tev utio ABr xfj utio AEZ, xf)v 8e utio BrA xfj utio EZA xal exi xrjv utio BAr xfj utio EAZ. Suveaxdxw yap Tipoc; xfj EZ eu-Ma xal xou; Tipoc; auxfj ot){ieioiz xou; E, Z xfj [lev utio ABr ywvia I'ar) f) utio ZEH, xfj 8s utio ArB for] f) utio EZH- XoiTtf] apa rj Tipoc; xw A Xomfj xfj Tipoc; tS H iaxiv I'ar). "Iaoycbvtov apa saxl xo ABr xpiywvov xw EHZ [xpiycbv- w] . xwv apa ABr, EHZ xpiycbvcov dvdXoyov eiaiv at TiXeupal at Tispl xdc; foac; yoviac; xal o^toXoyoi at utio xdc; foac; ywvtac; UTioxstvouaai- eaxiv apa (be; f] AB Tipoc; xrjv Br, [ouxwc;] f] HE Tipoc; xf]v EZ. dXX'' 6<; f] AB Tipoc; xf]v Br, ouxwc; UTioxsixai f) AE Tipoc; xf)v EZ- cbc; apa f) AE Tipoc; xf]v EZ, ouxox f] HE Tipoc; xf]v EZ. exaxepa apa xwv AE, HE Tipoc; xrjv EZ xov auxov s^si Xoyov Tar] apa saxtv f] AE xfj HE. 81a xd auxd 8f] xal f] AZ xfj HZ eaxiv \ar\. kizzi ouv I'ar) eaxlv f] AE xfj EH, xoivf] 8e f) EZ, 860 8f) at AE, EZ 8ual xdic; HE, EZ foai etaiv xal pdaic; f] AZ pdaei xfj ZH [eaxiv] for) - yovia apa f) utio AEZ ywvia xfj utio HEZ eaxiv for), xal xo AEZ xplywvov xcp HEZ xpiycbvo foov, xal at Xoiiiat yoviai xau; XoiTidic; ywvtaic foai, ucp' ac; at foai TiXeupal UTioxeivouaiv. for] apa laxl xal f) \iev utio AZE ya>via xfj utio HZE, f) 8e utio EAZ xfj utio EHZ. xal etc el f) [Lev utio ZEA xfj Otio HEZ eaxiv for), dXX' f) utio HEZ xfj utio ABr, xal f) Otio ing equal angles correspond. (Which is) the very thing it was required to show. Proposition 5 If two triangles have proportional sides then the trian- gles will be equiangular, and will have the angles which corresponding sides subtend equal. A D Let ABC and DEF be two triangles having propor- tional sides, (so that) as AB (is) to BC, so DE (is) to EF, and as BC (is) to CA, so EF (is) to FD, and, fur- ther, as BA (is) to AC, so ED (is) to DF. I say that triangle ABC is equiangular to triangle DEF, and (that the triangles) will have the angles which corresponding sides subtend equal. (That is), (angle) ABC (equal) to DEF, BCA to EFD, and, further, BAC to EDF. For let (angle) FEG, equal to angle ABC, and (an- gle) EFG, equal to ACB, have been constructed on the straight-line EF at the points E and F on it (respectively) [Prop. 1.23]. Thus, the remaining (angle) at A is equal to the remaining (angle) at G [Prop. 1.32]. Thus, triangle ABC is equiangular to [triangle] EGF. Thus, for triangles ABC and EGF, the sides about the equal angles are proportional, and (those) sides subtend- ing equal angles correspond [Prop. 6.4]. Thus, as AB is to BC, [so] GE (is) to EF. But, as AB (is) to BC, so, it was assumed, (is) DE to EF. Thus, as DE (is) to EF, so GE (is) to EF [Prop. 5.11]. Thus, DE and GE each have the same ratio to EF. Thus, DE is equal to GE [Prop. 5.9]. So, for the same (reasons), DF is also equal to GF. Therefore, since DE is equal to EG, and EF (is) common, the two (sides) DE, EF are equal to the two (sides) GE, EF (respectively). And base DF [is] equal to base FG. Thus, angle DEF is equal to angle GEF [Prop. 1.8], and triangle DEF (is) equal to triangle GEF, and the remaining angles (are) equal to the remaining angles which the equal sides subtend [Prop. 1.4]. Thus, angle DFE is also equal to GFE, and 161 ETOIXEIfiN 9'. ELEMENTS BOOK 6 ABr dpa ytovia xfj Otco AEZ eaxiv for). 81a xd auxd 8f] xal f] Otio ArB xfj Otio AZE eaxiv Tar), xal exi f) Tipog xo A xfj Tipoc; xw A- iaoycbviov apa eaxl xo ABr xpiycovov xfi AEZ xpiyci>vq>. 'Edv apa Buo xpiywva xag TiXeupdg dvdXoyov eyj), iaoycovia eaxai xd xpiywva xal laag e^ei xdg ycjviag, ucp' a<; al ojioXoyoi TtXeupal OTioxeivooaiv OTiep eSei Sel^ai. (angle) EDF to £G.F. And since (angle) FED is equal to GEF, and (angle) GEF to ^BC", angle ABC is thus also equal to DEF. So, for the same (reasons), (angle) ACB is also equal to DFE, and, further, the (angle) at A to the (angle) at D. Thus, triangle ABC is equiangular to triangle DEF. Thus, if two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal. (Which is) the very thing it was required to show. Proposition 6 'Edv 860 xpiytova ^uav ycoviav [iia ycovia forjv ey^r), Tiepl 8s xd<; foa<; ywviag xdc; TiXeupdg dvdXoyov, iaoycovia eaxai xd xpiytova xal laac, e$ei xd<; ytoviag, ucp' ac, ai ojioXoyoi TiXeupal UTioxeivouaiv. A A b r 'Eaxto 860 xpiyova xd ABr, AEZ jiiav ycoviav xfjv utio BAT [Lia ycovia xfj Otio EAZ forjv eyovxa, Tiepl Be xdc Xaac, ycoviac xac; TiXeupdg dvdXoyov, cb<; xf)v BA Tipoc; xfjv Ar, ooxcoc; xf]v EA npbz xf]V AZ- Xeyco, 6x1 iaoycoviov eaxi xo ABr xpiycovov xcp AEZ xpiy wvo xal iar]v e£ei xf)v Otio ABr ycoviav xfj hub AEZ, xfjv Be Otio ArB xfj bub AZE. Suveaxdxco yap Tipoc; xfj AZ eu-deia xal xoI<; Tipoc; ai) xfj at][ieioiz xolc; A, Z oTioxepa ^tev xcov bub BAT, EAZ for] f] Otio ZAH, xfj Be utio ArB Xar\ f) Otio AZH- XoiTif) apa f) Tipoc; xco B ycovia XoiTifj xfj Tipoc; iu H I'ar) eaxiv. 'Iaoycoviov apa eaxl xo ABr xpiycovov iu AHZ xpiycovcp. dvdXoyov apa eaxiv cbc; f] BA Tipoc; xfjv Ar, ouxcoc; f] HA Tcpoc; xfjv AZ. UTioxeixai Be xal cbc; f] BA Tipoc; xfjv Ar, ooxcoc; f] EA Tipoc; xfjv AZ- xal cbc; apa f] EA Tipoc; xfjv AZ, ooxcoc; f) HA Tipoc; xf]v AZ. for] apa f] EA xfj AH- xal xoivf] f] AZ- Buo Sf] ai EA, AZ 8ual xaTc; HA, AZ Xaac, eiaiv xal ycovia f) Otio EAZ ycovia xfj Otio HAZ [eaxiv] Xat]- pdaic; apa f] EZ pdaei xfj HZ eaxiv Xat], xal xo AEZ xpiycovov iS HAZ xpiycovcp I'aov eaxiv, xal ai Xoiical ycoviai xau; XoiTidic; ycoviaic; Xaac, eaovxai, ucp' ac; I'aag TiXeupal UTioxeivouaiv. for] apa eaxiv f) [lev utio AZH xfj Otio AZE, f] Be Otio AHZ If two triangles have one angle equal to one angle, and the sides about the equal angles proportional, then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal. B C Let ABC and DEF be two triangles having one angle, BAC, equal to one angle, EDF (respectively), and the sides about the equal angles proportional, (so that) as BA (is) to AC, so ED (is) to DF. I say that triangle ABC is equiangular to triangle DEF, and will have angle ABC equal to DEF, and (angle) ACB to DFE. For let (angle) FDG, equal to each of BAC and EDF, and (angle) DFG, equal to ACB, have been con- structed on the straight-line AF at the points D and F on it (respectively) [Prop. 1.23]. Thus, the remaining angle at B is equal to the remaining angle at G [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DGF. Thus, proportionally, as BA (is) to AC, so GD (is) to DF [Prop. 6.4]. And it was also assumed that as BA is) to AC, so ED (is) to DF. And, thus, as ED (is) to DF, so GD (is) to DF [Prop. 5.11]. Thus, ED (is) equal to DG [Prop. 5.9]. And DF (is) common. So, the two (sides) ED, DF are equal to the two (sides) GD, DF (respectively). And angle EDF [is] equal to angle GDF. Thus, base EF is equal to base GF, and triangle DEF is equal to triangle GDF, and the remaining angles 162 ETOIXEIfiN 9'. ELEMENTS BOOK 6 xfj utio AEZ. dXX' f] Otio AZH xfj utco ArB eoxw for) - xal f) utco ArB apa xfj utco AZE eaxiv lay). UTCoxsraa 8s xal #1 utco BAr xf) utco EAZ I'ar)' xal Xoitct) apa f] Tcp6<; iu B Xomfj xf) Tcpoc; xt5 E lar) eaxlv laoycoviov apa eaxl xo ABr xplywvov tu AEZ xpiytovtp. Eav apa 860 xptyova uiav yoviav [iia yovia larjv EXTl; Tcepl 8e xa<; laa? ytovlac; xa<; TcXeupac; dvdXoyov, iaoyiovia eaxai xa xpiywva xal laac, e^ei xa; ytav[a<;, ucp' ac; ai o^ioXoyoi nXeupal UTCoxeivouaiv oTcep eSei SeT^ai. Eav Buo xpiycova [da\ ytoviav (iia ywvia I'orjv evt), Tcepl 8e dXXac; yioviac; xdg TcXeupdc; dvdXoyov, xaiv 8e XoitcGv exaxepav aua f]xoi eXdaaova fj [j.r) eXdaaova opdrjc, iaoycbvia eaxai xa xpiycova xal laac, e<;ei xa<; y«via<;, Tcepl ac dvdXoyov eiaiv ai TtXeupai. 'Eaxw 860 xpiycova xa ABr, AEZ ^uav ycoviav [Lia ycovla iar]v e)(ovxa xfjv utco BAr xf] utco EAZ, Tcepl 8s dXXac; ycov(a<; xa<; utco ABr, AEZ xac; TcXeupdc; dvdXoyov, cog x/)v AB Tcpoc; xr)v Br, ouxcog xf)v AE Tcpoc; xf)v EZ, xcov 8e Xoitccov xcov Tcpoc; xolc; T, Z Tcpoxepov exaxepav d(ia eXdaaova op-drje Xeyco, oxi iaoycoviov eaxi xo ABr xpiycovov xco AEZ xpiycovcp, xal I'ar) eaxai f) utio ABr ycovia xrj utco AEZ, xal XoiTcfj BrjXovoxi f] Tcpoc; xcp T XoiTcfj xfj Tcpoc; xco Z Tar). EE yap dviaoc; eaxiv f) utio ABr ycovia xfj utio AEZ, fiia auxtov (lei^cov eaxiv. eaxw ^.ei^cov rj utco ABr. xal au- veaxdxco Tcpog xfj AB euiMa xal xco Tcpoc; auxfj arj^eico xcp B xrj utco AEZ ycovia Tar) rj utio ABH. Kal STtel Xar\ eaxiv f) [iev A ycovia xfj A, f) Be utco ABH xfj utco AEZ, Xoitct) apa f) utco AHB XoiTcfj xfj utco AZE eaxiv lay), iaoycoviov apa eaxl xo ABH xpiycovov ifi AEZ will be equal to the remaining angles which the equal sides subtend [Prop. 1.4]. Thus, (angle) DFG is equal to DFE, and (angle) DGF to DBF. But, (angle) DFG is equal to ACB. Thus, (angle) ACB is also equal to DFE. And (angle) BAG was also assumed (to be) equal to EDF. Thus, the remaining (angle) at B is equal to the remaining (angle) at E [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DEF. Thus, if two triangles have one angle equal to one angle, and the sides about the equal angles proportional, then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal. (Which is) the very thing it was required to show. Proposition 7 If two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles either both less than, or both not less than, right-angles, then the triangles will be equiangular, and will have the angles about which the sides are pro- portional equal. A Let ABC and DEF be two triangles having one an- gle, BAG, equal to one angle, EDF (respectively), and the sides about (some) other angles, ABC and DEF (re- spectively), proportional, (so that) as AB (is) to BC, so DE (is) to EF, and the remaining (angles) at C and F, first of all, both less than right-angles. I say that triangle ABC is equiangular to triangle DEF, and (that) angle ABC will be equal to DEF, and (that) the remaining (angle) at C (will be) manifestly equal to the remaining (angle) at F . For if angle ABC is not equal to (angle) DEF then one of them is greater. Let ABC be greater. And let (an- gle) ABG, equal to (angle) DEF, have been constructed on the straight-line AB at the point B on it [Prop. 1.23]. And since angle A is equal to (angle) D, and (angle) ABG to DEF, the remaining (angle) ACB is thus equal 163 ETOIXEIfiN 9'. ELEMENTS BOOK 6 xpiyovw. eaxiv dpa 6? f] AB Ttpoc; xrjv BH, ouxcoc; rj AE 7ip6<; xr]v EZ. &>q 8e f] AE Ttp6<; xf]v EZ, [ouxgk] unoxeixai f) AB Ttpoc x/]v Br- f] AB apa Ttpoc; exaxepav xwv Br, BH xov auxov exei Xoyov iar] apa f) Br xrj BH. waxe xai ycovia rj jcpo<; iS r ywvia xfj utco BHr eaxiv I'ar). eXdxxwv 8e op^rjc; Oicoxeixai r] Ttpoc; xc5 T- eXdxxov dpa eaxiv opi^c; xal utio BHr- waxe f] ecpe^rjc; auxfj yovia f) utco AHB [ie(£«v eaxiv op^rjc;. xal eSe()cdr] Iar] ouaa xfj Ttpoc; xw Z- xal f\ npbq x£> Z dpa [ie(Cwv eaxiv op'drjc;. UTioxeixai Be eXdaawv op'drjc;- OTtep eaxiv dxoKov. oux dpa dviaoc; eaxiv f) utco ABr ywvia xfj utco AEZ- iar) dpa. eaxi 8e xal f] Ttpoc; x£> A iar] xfj Ttpoc; tu A- xal Xoircf] apa r) Ttpoc; iw T XoiTtfj xfj Ttpoc; xw Z Tar] eaxiv. iaoywviov dpa eaxi xo ABr xpiywvov x£> AEZ xpiywvw. AXXa 5r) TtdXiv UTtoxeia'dw exaxepa xwv Ttpoc; xou; T, Z \xr\ eXdaaov op'drjc- Xeyco TtdXiv, oxi xal ouxoc eaxiv iaoyciviov xo ABr xpiywvov iS AEZ xpiywvw. Twv yap auxfiv xaxaaxeuaai!)evxcL)v 6[ioitL>c Be(c;o^ev, 6x1 Tar] eaxiv f) Br xrj BH- waxe xal ywvia f] npbq, xw T xfj utco BHr Tar] eaxiv. oux eXdxxov 8e op'dfjc r\ Ttpoc x£> E oux eXdxxcov apa op'drjc ouSe f) Otco BHr. xpiywvou 8r] xou BHr ai 8uo yoviai 860 op'dwv oux eiaiv eXdxxovec OTtep eaxiv dBuvaxov. oux dpa TidXiv dviaoc eaxiv f) utio ABr ywvia xfj utco AEZ- Tar] dpa. eaxi 8e xal f] npbq iw A xfj Ttpoc xw A Tar)- XoiTtr) apa f] Ttpoc xw T Xomfj xfj Ttpoc x<3 Z Tar] eaxiv. laoywviov apa eaxi xo ABr xpiywvov xw AEZ xpiywvw. 'Eav dpa 860 xpiyova jiiav ywviav (iia yovia iarjv e)(r), Ttepi Be dXXac ycoviac xac TtXeupdc dvdXoyov, xov Be XoiTtov exaxepav d^ia eXdxxova fj \±r\ eXdxxova op^fji;, iaoywvia eaxai xd xpiywva xallaac e<;ei iac, yoviac, nepl ac, dvdXoyov eiaiv ai uXeupai- oiiep e8ei 8el5ai- 'Eav ev 6pi3oy«v(w xpiywvcp aKO -zr\q op-dfj^ ycoviac sni x/]v pdaiv xdif)exo<; dx^fj, xd npbq xrj xa-dexcp xpiywva o^oid eaxi tu xe oXw xai aXkfikoiq. 'Eaxw xpiywvov 6pif)oywviov xo ABr 6p{>f]v e^ov x/]v UTio BAr ywviav, xal f]xi3w duo xou A era xr]v Br xd-dexoc; f] AA- Xeyw, oxi b[Loiov eaxiv exdxepov xwv ABA, AAr to the remaining (angle) DFE [Prop. 1.32]. Thus, trian- gle ABG is equiangular to triangle DEF. Thus, as AB is to BG, so DE (is) to EF [Prop. 6.4]. And as L>£ (is) to EF, [so] it was assumed (is) AB to BC. Thus, has the same ratio to each of BC and BG [Prop. 5.11]. Thus, BC (is) equal to BG [Prop. 5.9]. And, hence, the angle at C is equal to angle BGC [Prop. 1.5]. And the angle at C was assumed (to be) less than a right-angle. Thus, (angle) BGC is also less than a right-angle. Hence, the adjacent angle to it, AGB, is greater than a right-angle [Prop. 1.13]. And (AGB) was shown to be equal to the (angle) at F. Thus, the (angle) at F is also greater than a right-angle. But it was assumed (to be) less than a right- angle. The very thing is absurd. Thus, angle ABC is not unequal to (angle) DEF. Thus, (it is) equal. And the (angle) at A is also equal to the (angle) at D. And thus the remaining (angle) at C is equal to the remaining (an- gle) at F [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DEF. But, again, let each of the (angles) at C and F be assumed (to be) not less than a right-angle. I say, again, that triangle ABC is equiangular to triangle DEF in this case also. For, with the same construction, we can similarly show that BC is equal to BG. Hence, also, the angle at C is equal to (angle) BGC. And the (angle) at C (is) not less than a right-angle. Thus, BGC (is) not less than a right-angle either. So, in triangle BGC the (sum of) two angles is not less than two right-angles. The very thing is impossible [Prop. 1.17]. Thus, again, angle ABC is not unequal to DEF. Thus, (it is) equal. And the (an- gle) at A is also equal to the (angle) at D. Thus, the remaining (angle) at C is equal to the remaining (angle) at F [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle DEF . Thus, if two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles both less than, or both not less than, right-angles, then the triangles will be equiangular, and will have the angles about which the sides (are) propor- tional equal. (Which is) the very thing it was required to show. Proposition 8 If, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another. Let ABC be a right-angled triangle having the angle BAC a right-angle, and let AD have been drawn from 164 ETOIXEIfiN 9'. ELEMENTS BOOK 6 xpiyovov oXcp tw ABr xal exi dXXf|Xoi<;. A b a r Tkei yap ' or l ecttIv f] Otto BAr xfj Otto AAB- opi}/) yap exaxepa' xal xoivf] xcov 8uo xpiycjvwv xou xe ABr xal xou ABA f] Tipoc; xfi B, Xoittt] apa f] Otto ArB XoiTtfj xfj utto BAA eaxiv Tar)- laoycoviov apa eaxl xo ABr xpiyiovov xG ABA xpiywvcp. eaxiv apa tbc; f) Br UTtoxeivouaa xfjv op'drjv xou ABr xpiywvou Ttpoc; xrjv BA UTtoxeivouaav xr)V opdrjv xou ABA xpiywvou, ouxw<; auxf] f] AB UTtoxeivouaa xf)V 7ip6<; xw r ywviav xou ABr xpiywvou npoQ xf)V BA UTtoxeivouaav xf]v tarjv xf)v Otto BAA xou ABA xpiywvou, xal exi f] AT Ttpoc; xf)v AA UTtoxeivouaav xf]v Ttpoc; ifi B ywviav xoivrjv xQv Buo xpiywvwv. xo ABr dpa xpiywvov x£S ABA xpiywvw laoycoviov xe eaxi xal xd<; rcepl xac; Xaac, yoviac; TtXeupdc; dvdXoyov ex £l - o\io\.o\ a\xa [eaxl] xo ABr xpiyovov tu ABA xpiycovcp. o^toiwc; §f] 8ei^o[iev, oxi xal tu AAr xpiywvw 6[loi6m eaxi xo ABr xpiywvov exdxepov dpa iwv ABA, AAr [xpiywvwv] 6\ioi6v eaxiv oXw iS ABr. Aeya> 8rj, oxi xal aXXVjXou; eaxiv o^toia xd ABA, AAr xpiyova. 'Euel yap opiDf] f] utio BAA op'dfj xfj Otto AAr eaxiv tar), dXXd ^ir)v xal r] utto BAA xfj Ttpoc xCS T eBeix^r] Tar), xal XoiTtf] apa f] Ttpoc; x£5 B XoiTtfj xfj utio AAr eaxiv Tar)- laoycoviov apa eaxl xo ABA xplywvov x« AAr xpiyovo. eaxiv dpa cbc; f) BA xou ABA xpiywvou UTtoxeivouaa xf)V Otto BAA Tip6<; xf)v AA xou AAr xpiycovou UTtoxeivouaav xf)V Ttpoc; iS r Tarjv xfj Otto BAA, ouxwc; auxf] f) AA xou ABA xpiywvou UTtoxeivouaa xf]v Ttpoc; xw B ytoviav Ttpoc; xf)V Ar UTtoxeivouaav xfjv Otto AAr xou AAr xpiycivou Tarjv xfj Ttpoc; xw B, xal exi f] BA Ttpoc; xf)V Ar UTtoxeivouaai xdc; op-ddc;- o^toiov apa eaxl xo ABA xpiywvov iu AAr xpiywvw. 'Eav apa ev 6pi!)oywviw xpiyovo dTto xfj? opi^fj^ ywviac; era xf)V pdaiv xd^exoc; dx'dfj, xd Ttpoc; xfj xa$exo xpiycova opioid eaxi xw xe oXw xal dXXf|Xoic; [oTtep e5ei Bel^ai]- A, perpendicular to BC [Prop. 1.12]. I say that triangles ABD and ADC are each similar to the whole (triangle) ABC and, further, to one another. A B DC For since (angle) BAG is equal to ADB — for each (are) right-angles — and the (angle) at B (is) common to the two triangles ABC and ABD, the remaining (an- gle) ACB is thus equal to the remaining (angle) BAD [Prop. 1.32]. Thus, triangle ABC is equiangular to tri- angle ABD. Thus, as BC, subtending the right-angle in triangle ABC, is to BA, subtending the right-angle in tri- angle ABD, so the same AB, subtending the angle at C in triangle ABC, (is) to BD, subtending the equal (an- gle) BAD in triangle ABD, and, further, (so is) AC to AD, (both) subtending the angle at B common to the two triangles [Prop. 6.4]. Thus, triangle ABC is equian- gular to triangle ABD, and has the sides about the equal angles proportional. Thus, triangle ABC [is] similar to triangle ABD [Dei. 6.1]. So, similarly, we can show that triangle ABC is also similar to triangle ADC. Thus, [tri- angles] ABD and ADC are each similar to the whole (triangle) ABC. So I say that triangles ABD and ADC are also similar to one another. For since the right-angle BDA is equal to the right- angle ADC, and, indeed, (angle) BAD was also shown (to be) equal to the (angle) at C, thus the remaining (an- gle) at B is also equal to the remaining (angle) DAC [Prop. 1.32]. Thus, triangle ABD is equiangular to trian- gle ADC. Thus, as BD, subtending (angle) BAD in tri- angle ABD, is to DA, subtending the (angle) at C in tri- angle ADC, (which is) equal to (angle) BAD, so (is) the same AD, subtending the angle at B in triangle ABD, to DC, subtending (angle) DAC in triangle ADC, (which is) equal to the (angle) at B, and, further, (so is) BA to AC, (each) subtending right-angles [Prop. 6.4]. Thus, triangle ABD is similar to triangle ADC [Def. 6.1]. Thus, if, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base 165 ETOIXEIfiN 9'. ELEMENTS BOOK 6 U6pio\j.a. 'Ex 8r) xouxou cpavepov, oxi edv £v opTJoyMviw xpiytovw duo xfj<; op-df^ y«vdi<; etii xr)v pdaic; xd-fJexoc; d^rj, f] dx'fMaa x£Sv xfjc; pdaeox x^ir^dxtov \±scr} dvdXoyov saxiv ojiep sSei BsT^ai. t In other words, the perpendicular is the geometric mean of the pieces. then the triangles around the perpendicular are similar to the whole (triangle), and to one another. [(Which is) the very thing it was required to show.] Corollary So (it is) clear, from this, that if, in a right-angled tri- angle, a (straight-line) is drawn from the right-angle per- pendicular to the base then the (straight-line so) drawn is in mean proportion to the pieces of the base.t (Which is) the very thing it was required to show 0'. Trie Scddarjt; eO'ddae; xo Ttpoaxax'dev uepo<; dcpeXeiv. A Z B TEaxco f] So-dsiaa eO-deia f] AB- BsT 8f) xfje AB xo Ttpo- axaxiJEV [izpoz dcpeXeiv. 'EKixsxdx'dw 8r] xo xpixov. [xdi] SiVj'dx 61 Tt< ^ TO ° A sMeia f\ AT ywviav izepie^ovaoi (iexd xfje AB xuxouaav xod slXr]Cpi&6j xuxov arj^islov em xfje Ar xo A, xdi xeiaOcoaav xfj AA I'aai al AE, EI\ xdi stte^sux'v}" #j Br, xdi 8id xoO A TtapdXXrjXoc; auxfj f^x'&w T) AZ. 'End ouv xpiycovou xou ABr napd [iiav xfiv TtXeupfiv xf]v Br rjxxoa rj ZA, dvdXoyov dpa eaxlv &>z f\ TA 7tp6<; xf]V AA, ouxcoc; f] BZ itpoe xrjv ZA. SiTiXfj 8e #) IA xfje AA- SiTtXrj dpa xdi f\ BZ xrjc; ZA- xpmXrj dpa f) BA xfje AZ. Tfjc; dpa Bo-deiaTr]? sO'deia? xfj? AB xo emxax'dev xpixov ^tepot; dcp/]pr]xai xo AZ- oitep s8ei Ttoifjaai. Proposition 9 To cut off a prescribed part from a given straight-line. A F B Let AB be the given straight-line. So it is required to cut off a prescribed part from AB. So let a third (part) have been prescribed. [And] let some straight-line AC have been drawn from (point) A, encompassing a random angle with AB. And let a ran- dom point D have been taken on AC. And let DE and EC be made equal to AD [Prop. 1.3]. And let BC have been joined. And let DF have been drawn through D parallel to it [Prop. 1.31]. Therefore, since FD has been drawn parallel to one of the sides, BC, of triangle ABC, then, proportionally, as CD is to DA, so BE (is) to FA [Prop. 6.2]. And CD (is) double DA. Thus, BF (is) also double FA. Thus, BA (is) triple AF. Thus, the prescribed third part, AF, has been cut off from the given straight-line, AB. (Which is) the very thing it was required to do. i'. Proposition 10 Trjv 8o$£iaav eCcdeiav dx|ir)xov xrj 8oi!)e[ar| xexur^evn To cut a given uncut straight-line similarly to a given b[LoioK xe^iav. cut (straight-line). 166 ETOIXEIfiN 9'. ELEMENTS BOOK 6 r A Z H B "Eaxw f] ^.ev So-daaa sCWteTcx ax[ir]xo<; f] AB, f] 8e xe- x(i/)[j.£vr) f) Ar xaxa xa A, E ar)U£la, xal xeicrdojaav &axe ywviav xuxouaav nepiexeiv, xal eto^sux'&m rj TB, xal Bid xfiv A, E xfj Br TiapdXXrjXoi fj/iJioaav ai AZ, EH, Sid 8e xou A xfj AB TtapdXXiqXoc; fj)(#6j f) A0K. napaXXr]X6ypa[j.|iov dpa eaxlv sxdxspov x«v ZO, OB- Tor) dpa f) [isv A6 xfj ZH, f] 8s 6K xfj HB. xal ercel xpiyiovou xou AKr napd jiiav iwv nXsupwv xrjv Kr euifeTa rjxxai f] 9E, dvdXoyov dpa Eaxlv foe, f) TE upoc; xrjv EA, ouxgjc; f) K6 Tipbc, x/)v 6A. lor) 8e f) ^ev K6 xfj BH, f] 8e OA xfj HZ. eaxiv dpa cb<; f] TE npoc, xrjv EA, ouxw<; f) BH npoc, xrjv HZ. TidXiv, ETtd xpiywvou xou AHE Ttapd |i(av xt5v TtXsupwv x/]v HE rjxxai t\ ZA, dvdXoyov dpa eaxlv 6<; r] EA 7tp6<; xrjv AA, ouxw<; f] HZ npoz xrjv ZA. eSdx'dr) 8e xal (be r) EE npoc; xr]v EA, ouxo<; f) BH npoc, xrjv HZ- eaxiv dpa «<; jiev f) TE npoc; xrjv EA, ouxcx fj BH npoc; xrjv HZ, cb<; 8e rj EA npoc xrjv AA, ouxck fj HZ 7tp6<; xrjv ZA. 'H dpa So'deTaa su^eTa dx^trjxoc; rj AB xfj So-Marj eu-Ma xexjurj^evyj xfj Ar b[io'ux>z xexurjxai- omp s8ei Ttoirjaar la'. Auo So-deiafiv eu'deifiv xpixrjv dvdXoyov npoasupeTv. 'Eaxwaav ai 8oif)eIaai [8uo eu'dETai] ai BA, Ar xal xsicrdtoaav ywvlav Ttepiixouaai xuxouaav. Bel 8rj xGv BA, Ar xpixrjv dvdXoyov upoaeupav. expepXfjcrdwaav yap em xd A, E arpeia, xal xslcxdo xfj Ar larj fj BA, xal e^eZ^X^ f] Br, xal Bid xou A rcapdXXrjXoc; auxfj fix&oi f] AE. Tkel ouv xpiycovou xou AAE Ttapd piiav xfiv nXeupfiv xrjv AE rjxxai fj Br, dvdXoyov egxiv &<; rj AB Ttpoc; xrjv BA, ouxwe; f) Ar Ttpoc; xrjv TE. Tar) Be f] BA xfj Ar. eoxiv dpa cb<; fj AB npog xrjv Ar, oux«<; rj Ar Ttpoc; xrjv TE. C A F G B Let AB be the given uncut straight-line, and AC a (straight-line) cut at points D and E, and let (AC) be laid down so as to encompass a random angle (with AB) . And let CB have been joined. And let DF and EG have been drawn through (points) D and E (respectively), parallel to BC, and let DHK have been drawn through (point) D, parallel to AB [Prop. 1.31]. Thus, FH and HB are each parallelograms. Thus, DH (is) equal to FG, and iJif to GB [Prop. 1.34]. And since the straight-line HE has been drawn parallel to one of the sides, KC, of triangle DKC, thus, proportionally, as CE is to ED, so KH (is) to #L> [Prop. 6.2]. And ifi? (is) equal to BG, and #L> to GF. Thus, as C£ is to ED, so £?G (is) to GF. Again, since FD has been drawn parallel to one of the sides, GE, of triangle AGE, thus, proportionally, as ED is to DA, so GF (is) to FA [Prop. 6.2]. And it was also shown that as CE (is) to ED, so SG (is) to GF. Thus, as CE is to £L>, so BG (is) to GF, and as ED (is) to FM, so GF (is) to FA. Thus, the given uncut straight-line, AB, has been cut similarly to the given cut straight-line, AC. (Which is) the very thing it was required to do. Proposition 11 To find a third (straight-line) proportional to two given straight-lines. Let BA and AC be the [two] given [straight-lines], and let them be laid down encompassing a random angle. So it is required to find a third (straight-line) proportional to BA and AC. For let {BA and AC) have been produced to points D and F (respectively), and let BD be made equal to AC [Prop. 1.3]. And let BC have been joined. And let DE have been drawn through (point) D parallel to it [Prop. 1.31]. Therefore, since BC has been drawn parallel to one of the sides FF of triangle ADE, proportionally, as AB is to BD, so AC (is) to GF [Prop. 6.2]. And BD (is) equal 167 ETOIXEIfiN 9'. ELEMENTS BOOK 6 A Auo apa Bo^Eiawv EuiDeiwv xwv AB, Ar xplxT] dvdAoyov auxaac; 7tpoaeupT]xai f] TE- onep eSei uoifjoai. LP'. Tpiwv Soifteiacov sMeicov xsxdpxrjv dvdXoyov Ttpo- aeupeTv. A' ' B ' r A z "Eaxwaav ai So'der'iaai xpstc; euiDeTai ad A, B, F- SeT 8t) xwv A, B, r xexpdxrjv dvdXoyov ixpooeupslv. 'ExxeiaiSwoav 860 eu'dslai ai AE, AZ ycoviav Tispis/oug- ai [xuxouaav] xt]v Otto EAZ- xal xsio-fko xfj Alar] r) AH, xf) 8s B laT) f) HE, xal exi xfj F far) f) A©- xal emZ^X^ e ' lar ]^ tt)c, HO KapdXXT)Xo<; auxfj ryyj}^ 81a xou E f) EZ. 'EtcsI odv xpiyiovou xou AEZ rcapa jjiav xtjv EZ T)xxai f] HO, eaxiv apa tlx; f) AH npbc, xf)v HE, ouxioc; f) AO Ttpoc; xt)v OZ. laT) 8e f) piev AH xt) A, f) 8e HE xt) B, f) 8s AO xt) T- eaxiv apa &>z r) A upog xt]v B, ouxgj<; f) T npbc, xtjv OZ. Tpicov apa Scdeiawv eu$eic5v x«v A, B, T xexdpxT) dvdXoyov iipoosupTjxai f] OZ- onep e8ei ixoifjaaL. to AC. Thus, as is to AC, so AC (is) to CE. A D E Thus, a third (straight-line), C_E, has been found (which is) proportional to the two given straight-lines, AB and AC. (Which is) the very thing it was required to do. Proposition 12 To find a fourth (straight-line) proportional to three given straight-lines. A' 1 B' 1 D H F Let A, B, and C be the three given straight-lines. So it is required to find a fourth (straight-line) proportional to A, B, and C. Let the two straight-lines DE and DF be set out en- compassing the [random] angle EDF. And let DG be made equal to A, and GE to B, and, further, DH to C [Prop. 1.3]. And GH being joined, let EF have been drawn through (point) E parallel to it [Prop. 1.31]. Therefore, since GH has been drawn parallel to one of the sides EF of triangle DEF, thus as DG is to GE, so DH (is) to HF [Prop. 6.2]. And DG (is) equal to A, and GE to B, and DH to C. Thus, as A is to £?, so C (is) 168 ETOIXEIfiN 9'. ELEMENTS BOOK 6 Auo Bo-dsiawv eutJeiGv ^isaiqv dvdXoyov TtpoasupsTv. a Br "Eaxwaav ad BoiDeToai 8uo eu'dsTai ai AB, Br- Bel 8r) xSv AB, Br jiearjv dvdXoyov TtpoaEupsTv. Keicrdtoaav etc' eu'fkiac;, xal ycypdcp'ScL) etu xrjc; Ar f]^uxuxXiov to AAr, xal t^x'&co duo xoO B ar^dou xfj Ar su-Ma Ttpoc; opMc; f] BA, xal ETce^etix'fltoaav ai AA, Ar. 'Etcei ev fj^ixuxXicp ycovia laxiv 7] Otco AAr, op'dr] eaxiv. xal etcei sv dpiJoycoviw xpiytovw xG AAr dico xrjc; opiDfjc; yoviac; era xr)v pdaiv xdiSsxoc rjxxai f\ AB, f] AB apa iSv xfj? pdoewi; x^ir^dxcov x£>v AB, Br \iear\ dvdXoyov ioxiv. Auo apa Bo'deiawv eu'deifiv xwv AB, Br \±£or} dvdXoyov icpoaeuprjxai f] AB- OTtep e8ei noifjaai. t In other words, to find the geometric mean of two given straight-lines. i5'. Toiv i'omv xs xal I'ooywviwv itapaXXr]Xoypdjj.(icov dvxi- Tteitovdaaiv ai icXsupal ai Ttepi xac; laac, ytoviac;- xal £>v ioo- ycoviwv iiapaXXrjXoypd^wv avxiTiETtovdaaiv ai TtXeupai ai Ttspl xac; laac; yoviac;, laa saxlv exeTva. "Eaxa> Taa xs xal iaoywvia TcapaXX/]X6ypa[i^.a xa AB, Br XaaLC, zyovxaL xac; Ttpoc; iu B ywviac;, xal xeicrdwaav etc' eMeiac; ai AB, BE- etc' cO^eiac; apa eiai xal ai ZB, BH. Xeyo, oxi xQv AB, Br avxiTieicovdaoiv ai TtXsupal ai Ttepi xac; laac; ywviac;, xouxeaxiv, oxi Eaxlv «<; f] AB Ttpoc; xr]v BE, ouxoc; f) HB Ttpoc; xrjv BZ. to HF. Thus, a fourth (straight-line), HF, has been found (which is) proportional to the three given straight-lines, A, B, and C. (Which is) the very thing it was required to do. Proposition 13 To find the (straight-line) in mean proportion to two given straight-lines.^ A B C Let AB and BC be the two given straight-lines. So it is required to find the (straight-line) in mean proportion to AB and BC. Let {AB and BC) be laid down straight-on (with re- spect to one another), and let the semi-circle ADC have been drawn on AC [Prop. 1.10]. And let BD have been drawn from (point) B, at right-angles to AC [Prop. 1.11]. And let AD and DC have been joined. And since ADC is an angle in a semi-circle, it is a right-angle [Prop. 3.31]. And since, in the right-angled triangle ADC, the (straight-line) DB has been drawn from the right-angle perpendicular to the base, DB is thus the mean proportional to the pieces of the base, AB and BC [Prop. 6.8 corr.]. Thus, DB has been found (which is) in mean propor- tion to the two given straight-lines, AB and BC. (Which is) the very thing it was required to do. Proposition 14 In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. And those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal. Let AB and BC be equal and equiangular parallelo- grams having the angles at B equal. And let DB and BE be laid down straight-on (with respect to one another). Thus, FB and BC are also straight-on (with respect to one another) [Prop. 1.14]. I say that the sides of AB and 169 ETOIXEIfiN 9'. ELEMENTS BOOK 6 e r A A Su^TtSTtXrjpcoa'dto yap to ZE TtapaXXr)X6ypa^i|iov. stisI ouv laov sera to AB TtapaXX/]X6ypa^ov iw Br TtapaXX/]- Xoypd^cp, aXXo Ss ti to ZE, sotiv apa (be; to AB Ttpoc; to ZE, outwc; to Br Ttpoc; to ZE. dXX'' (be; [Lev to AB upoc; to ZE, outwc; f] AB Ttpoc; ttjv BE, (be; 8s to Br Ttpoc; to ZE, outw<; f] HB Ttpoc; t/]v BZ- xal (be; apa f] AB Ttpoc; t/]v BE, outoc; f\ HB Ttpoc; ttjv BZ. twv apa AB, Br TtapaX- Xf]koyp6i[i[Ux>\i dvTiTtSTtovdaaiv al TtXsupal al Ttspl Tac; 1'aac; yovlac;. AXXd 5f] screw (be. f] AB Ttpoc; ttjv BE, outcoc; f\ HB Ttpoc; t/]v BZ- Xsyw, oil laov feaxl to AB TtapaXXr)X6ypa^ov iu Br TtapaXXr)Xoypd^«. Tksl yap sotiv to? #] AB Ttpoc; ttjv BE, outwc; #] HB Ttpoc; tt)v BZ, dXX'' (be; ^isv rj AB Ttpoc; ttjv BE, outcoc; to AB TtapaXX/]X6ypa^ov Ttpoc; to ZE TtapaXXr)X6ypa^ov, (be; 8s f] HB Ttpoc; ttjv BZ, outmc; to Br TtapaXX/]X6ypa^ov Ttpoc; to ZE TtapaXX/]X6ypa^ov, xal (be; apa to AB Ttpoc; to ZE, oOtwc; to Br Tipoc; to ZE- laov apa taxi to AB TtapaXX/]X6ypa^ov tco Br TtapaXX/]Xoypdu[jicp. TGv apa 'lacov ts xal laoyovlcov TtapaXXrjXoypd^wv avTiTCSTiovdacuv al TtXsupal al Ttspl Tac; 1'aac; ywvlac;- xal SSv laoyovlov TtapaXXrjXoypd^cov dvTiTtSTtovdaaiv al TtXsupal al icspl Tac; 'laac; ywvlac;, I'aa scrav sxslva- OTtsp sBsi 8sTc;ai. is'. Tcov iacov xal ^uav ^tia \ar\v sxovtcov ywvlav Tpiycbvcov dvTiTtSTtovdaaiv al TtXsupal al Ttspl Tac; laac; ycovlac;- xal £>v [ilav (iia I'oiqv s)(6vt(ov ycovlav Tpiytbvtov avTiTtSTtovdaaiv al TtXsupal al Ttspl Tac; 'laac; ycovlac;, laa sotIv sxslva. 'Earco !aa Tplycova to ABr, AAE jilav [Lia iar)v sxovTa ycovlav ttjv uno BAr Tfj utio AAE- Xsyto, oti tGv ABr, AAE Tpiyovwv dvTiTtSTtovdaaiv al TtXsupal al Ttspl Tac; i'aac; ycovlac;, toutsctuv, oti sgtIv (be; i] FA Ttpoc; t/]v AA, outcoc; BC about the equal angles are reciprocally proportional, that is to say, that as DB is to BE, so GB (is) to BF. E C A D For let the parallelogram FE have been completed. Therefore, since parallelogram AB is equal to parallelo- gram BC, and FE (is) some other (parallelogram), thus as (parallelogram) AB is to FE, so (parallelogram) BC (is) to FE [Prop. 5.7]. But, as (parallelogram) AB (is) to FE, so DB (is) to BE, and as (parallelogram) BC (is) to BB, so GB (is) to [Prop. 6.1]. Thus, also, as DB (is) to BE, so GB (is) to BF. Thus, in parallelograms AB and BC the sides about the equal angles are reciprocally proportional. And so, let DB be to BE, as GB (is) to BF. I say that parallelogram AB is equal to parallelogram BC. For since as DB is to BE, so GB (is) to BF, but as BB (is) to BE, so parallelogram AB (is) to parallelo- gram FE, and as GB (is) to BF, so parallelogram BC (is) to parallelogram FE [Prop. 6.1], thus, also, as (par- allelogram) AB (is) to FE, so (parallelogram) BC (is) to FE [Prop. 5.11]. Thus, parallelogram AB is equal to parallelogram BC [Prop. 5.9]. Thus, in equal and equiangular parallelograms the sides about the equal angles are reciprocally propor- tional. And those equiangular parallelograms in which the sides about the equal angles are reciprocally propor- tional are equal. (Which is) the very thing it was required to show. Proposition 15 In equal triangles also having one angle equal to one (angle) the sides about the equal angles are reciprocally proportional. And those triangles having one angle equal to one angle for which the sides about the equal angles (are) reciprocally proportional are equal. Let ABC and ADE be equal triangles having one an- gle equal to one (angle), (namely) BAC (equal) to DAE. I say that, in triangles ABC and ADE, the sides about the 170 ETOIXEIfiN 9'. ELEMENTS BOOK 6 f] EA Ttpoc x/]v AB. b r A E Ksia'dw yap uoxe en su-deiac slvai xrjv EA xfj AA- eti' eu-Qeiac, apa eaxl xal f] EA xrj AB. xal ETteCeux'dw f] BA. Ekei ouv laov taxi xo ABr xpiycovov xw AAE xpiywvw, aXXo 8e xi xo BAA, saxiv apa cbc xo TAB xpiywvov Ttpoc xo BAA xpiycovov, ouxwc xo EAA xpiywvov Ttpoc xo BAA xpiyovov. dXX' cbc ^.ev xo TAB Ttpoc xo BAA, ouxwc f] TA Ttpoc x/]v AA, 6? Ss xo EAA Ttpoc xo BAA, ouxtoc f] EA Ttpoc x/]v AB. xal <i>c apa f] TA Ttpoc x/]v AA, ouxoc f] EA Ttpoc x/]v AB. xwv ABr, AAE apa xpiywvwv dvxi- TiETtovdaaiv ai TtXsupal al Ttspl xdc taac ycoviac. AXXa 8r) avxiKETiovdsxcoaav ai TtXeupal xwv ABr, AAE xpiycivov, xal saxco wc f] EA Ttpoc xrjv AA, ouxmc f] EA Tipoc xr]v AB- Xsyw, oxi taov eaxl xo ABr xpiywvov iw AAE xpiyova). 'Emc'sux'deiaTjc yap TtdXiv xrjc BA, stcei eaxiv «c f] TA Ttpoc; xf]v AA, ouxwc r) EA Ttpoc xrjv AB, dXX'' cbc jiev f\ TA Tipoc x/]v AA, ouxcoc xo ABr xplywvov Ttpoc xo BAA xpiywvov, cbc 5e rj EA Ttpoc xrjv AB, ouxcoc xo EAA xpiyovov Ttpoc xo BAA xpiywvov, (be apa xo ABr xplywvov Ttpoc xo BAA xpiywvov, oux«c xo EAA xpiywvov Ttpoc xo BAA xpiywvov. exdxepov apa xwv ABr, EAA Ttpoc xo BAA xov auxov Xoyov. lacov apa eaxl xo ABr [xpiycovov] xo EAA xpiyovco. Twv apa lawv xal ^.[av \ua larjv £)(6vxcl>v ywvlav xpiycbvov dvxiTtSTtovdaaiv ai TtXeupal ai Ttepl xdc I'aac ycoviac xal 5c ^iav (iia larjv e)(6vx«v ywviav xpiyovov avxiTteitov^aaiv ai TtXeupal ai Ttepl xdc laac, yoviac, exelva taa eaxlv oTtep eSsi 5eTcai. 'Edv xeaaapec euiDelai dvdXoyov Saiv, xo utio xfiv dxptov Tt£pi£)(6|jievov op-doytbviov i'aov Eaxl ifi utio xfiv (jifawv Tiepie/o^evo opiJoywviw- xav xo utio xGv axpwv equal angles are reciprocally proportional, that is to say, that as CA is to AD, so EA (is) to AB. B C D E For let CA be laid down so as to be straight-on (with respect) to AD. Thus, EA is also straight-on (with re- spect) to AB [Prop. 1.14]. And let BD have been joined. Therefore, since triangle ABC is equal to triangle ADE, and BAD (is) some other (triangle), thus as tri- angle CAB is to triangle BAD, so triangle EAD (is) to triangle BAD [Prop. 5.7]. But, as (triangle) CAB (is) to BAD, so CA (is) to AD, and as (triangle) EAD (is) to BAD, so EA (is) to AB [Prop. 6.1]. And thus, as CA (is) to AD, so EA (is) to Ai?. Thus, in triangles ABC and ADE the sides about the equal angles (are) reciprocally proportional. And so, let the sides of triangles ABC and ADE be reciprocally proportional, and (thus) let C A be to AD, as EA (is) to AB. I say that triangle ABC is equal to triangle ADE. For, BD again being joined, since as CA is to AD, so £A (is) to AB, but as CA (is) to AD, so triangle ABC (is) to triangle BAD, and as _EA (is) to AB, so triangle £A£> (is) to triangle BAD [Prop. 6.1], thus as triangle ABC (is) to triangle BAD, so triangle EAD (is) to tri- angle BAD. Thus, (triangles) ABC and EAD each have the same ratio to BAD. Thus, [triangle] ABC is equal to triangle EAD [Prop. 5.9]. Thus, in equal triangles also having one angle equal to one (angle) the sides about the equal angles (are) recip- rocally proportional. And those triangles having one an- gle equal to one angle for which the sides about the equal angles (are) reciprocally proportional are equal. (Which is) the very thing it was required to show. Proposition 16 If four straight-lines are proportional then the rect- angle contained by the (two) outermost is equal to the rectangle contained by the middle (two) . And if the rect- 171 ETOIXEIfiN 9'. ELEMENTS BOOK 6 Ti£pi£)(6^£vov op'doytoviov loov rj tw (mo xcov (isawv Tiepie- /o|ji£vw op-doytoviw, ai xeaaapec; eO'deTai dvaXoyov eaovxai. H A e b r angle contained by the (two) outermost is equal to the rectangle contained by the middle (two) then the four straight-lines will be proportional. H B C D E' ' "Ecraoaocv xeaaapec; eu'delai dvaXoyov ai AB, TA, E, Z, cbc; f) AB Tipoc; tt)v TA, ouxgjc; f) E Tipoc; xf)v Z' Xeyio, oxi xo utio xcov AB, Z 7iepi£)(6[i£vov op'doywviov laov eaxi tu uno xov TA, E Tiepiexo^tevw opTJoyMviw. "Hx^waav [yap] duo xwv A, T ar^eitov xdic; AB, TA eu-deiaic; Tipoc; op-ddc; ai AH, TO, xai xeiaiDw xfj (jiev Z tar) f] AH, xfj Se E i'ar] f] TO. xai auuTieTiXir]pt5a , d« xd BH, A6 TiapaXX/]X6ypa^a. Kai CTiei eaxiv w<; f] AB Tipoc; xrjv TA, ouxwc; f] E Ttpoc xf)v Z, iar) 5e f] ^tev E xfj r0, f] 8e Z xfj AH, eaxiv dpa w<; f] AB Tipoc; xf)v TA, ouxck f] T0 Tipoc; xf)v AH. xfiv BH, A0 apa TiapaXXrjXoypdu^iMv dvxiTiCTiovdaaiv ai TiXeupal ai Tcspl xac; laac; yoviac;. Sv 8e iaoytoviwv TiapaXXr)Xoypd^(jiov dvxiTienovdaaiv ai TiXeupai ai Tiepi xac; i'aac; ytovdic;, laa eaxiv exelva- laov apa eaxi xo BH TiapaXX/]X6ypa^ov x<5 A6 7iapaXX/]Xoypd^^.CL>. xai eaxi xo jiev BH xo uno xCSv AB, Z' iar] yap f] AH xfj Z- xo Se A0 xo utio xfiv TA, E- Xar] yap f) E xfj T9- xo apa utio twv AB, Z Tiepiexo^evov op'doytoviov laov eaxi x£5 uno xwv TA, E Txepiexo^evw op-doytoviw. AXXd 8f] xo utio xfiv AB, Z Tiepiexo^tevov op-doycoviov laov eaxw xw utio twv TA, E Tiepiexo^tevo 6pi9oycovicp. Xeyo, oxi ai xeaaapec eu'delai dvaXoyov eaovxai, dbc; f] AB Tipoc; xf)v TA, ouxoc; f) E Tipoc xf)V Z. Tcov yap auxfiv xaxaaxeuaa-devxcov, etxel xo utio twv AB, Z laov eaxi xG utio iuv TA, E, xai eaxi xo ^.ev utio x£Sv AB, Z xo BH- Iar) yap eaxiv f) AH xfj Z- xo 8e utio icov TA, E xo A9- i'ar) yap f) T8 xfj E- xo apa BH i'aov eaxi tu A0. xai eaxiv iaoywvia. iuv 8e lawv xai iaoyovicov TiapaXX/jXoypd^tov dvxiTienovdaaiv ai TiXeupal ai Tiepi xac; laac; ywviac;. eaxiv apa &>z f] AB Tipoc; xf]v TA, ouxoc f] T& Tipoc; xf)V AH. iar] 8e f) ^tev r0 xfj E, f) 8e AH xfj Z- eaxiv apa cbc; f] AB Tipoc; xf)V TA, ouxw<; f\ E Tipoc xfjv Z. 'Edv apa xeaaapec euiJeTai dvaXoyov waiv, xo utio xwv axpwv nepiexo^ievov op-doycoviov i'aov eaxi xw utio tuv ^ifawv nepiexo^ievw op-doyMviw- xav xo utio xwv axpwv Tiepiex6[ievov op-doywviov laov fj xQ utio xwv [icawv iiepie- Xo^tevw op'doywvico, ai xeaaapec eu^elai dvaXoyov eaovxar oTiep e8ei 8eT^ai. Ei 1 F' 1 Let AB, CD, E, and F be four proportional straight- lines, (such that) as AB (is) to CD, so E (is) to F. I say that the rectangle contained by AB and F is equal to the rectangle contained by CD and E. [For] let AG and CH have been drawn from points A and C at right-angles to the straight-lines AB and CD (respectively) [Prop. 1.11]. And let AG be made equal to F, and CH to E [Prop. 1.3]. And let the parallelograms BG and DH have been completed. And since as AB is to CD, so E (is) to F, and E (is) equal CH, and F to AG, thus as AB is to CD, so C# (is) to AG. Thus, in the parallelograms BG and Di? the sides about the equal angles are reciprocally propor- tional. And those equiangular parallelograms in which the sides about the equal angles are reciprocally propor- tional are equal [Prop. 6.14]. Thus, parallelogram BG is equal to parallelogram DH. And BG is the (rectangle contained) by AB and F. For AG (is) equal to F. And DH (is) the (rectangle contained) by CD and E. For E (is) equal to CH. Thus, the rectangle contained by AB and F is equal to the rectangle contained by CD and E. And so, let the rectangle contained by AB and F be equal to the rectangle contained by CD and E. I say that the four straight-lines will be proportional, (so that) as AB (is) to CD, so E (is) to F. For, with the same construction, since the (rectangle contained) by AB and F is equal to the (rectangle con- tained) by CD and E. And BG is the (rectangle con- tained) by AB and F. For AG is equal to F. And DH (is) the (rectangle contained) by CD and E. For CiJ (is) equal to E. BG is thus equal to DH. And they are equiangular. And in equal and equiangular parallel- ograms the sides about the equal angles are reciprocally proportional [Prop. 6.14]. Thus, as AB is to CD, so CH (is) to AG. And CH (is) equal to E, and AG to F. Thus, as AB is to CD, so E (is) to F. Thus, if four straight-lines are proportional then the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two) . And if the rectangle contained by the (two) outermost is equal to 172 ETOIXEIfiN 9'. ELEMENTS BOOK 6 'Edv xpeu; eO'delai dvdXoyov Gmv, to Oko xGv dxptov Tiepiexo^ievov op-doyGviov laov eaxl xG &7t6 xfjc; near)? xe- xpayGvt>y xav xo utio xGv dxptov Tiepie/ojjievov op'doyGviov I'aov f) xG duo xfjc; ^earjc; xexpayGvcp, ai xpeu; eiWteTai dvdXoyov eaovxai. A' 1 B' 1 A i 1 r> ' "Eaxwaav xpeu; eu-deTai dvdXoyov ai A, B, r, Gc; f] A Tipoc; xf]v B, ouxtoc; rj B Tipoc; xf]v E Xeyco, oxi xo Oreo xGv A, r Tiepie)(6^evov op-doyGviov I'aov eaxl xG diio xfjc; B xexpayGvtp. Eeia-dco xfj B iar] f] A. Kdi ercei eaxiv Gc; f] A Tipoc; xf|V B, ouxw<; f] B Tipoc; xf]v T, iar) 8e f] B xfj A, eaxiv dpa Gc; f] A Tipoc; xrjv B, f] A Tipoc; xf|v r. edv 8e xeaaapec; euiDeTai dvdXoyov Gaiv, xo utio twv dxpwv Tiepie)(6^evov [opiJoyGviov] i'aov eaxl xo uko xGv (irotov Tiepiexo^iEvo opiJoyoviw. xo dpa utio tuv A, T iaov eaxl xG utio xGv B, A. dXXd xo utio xGv B, A xo duo xfjc; B eaxiv iar] ydp r\ B xfj A- xo dpa utio twv A, T Tiepie)(6[ievov opiJoyGviov laov eaxl xo duo xfjc; B xexpayGvw. AXXd 8f] xo utio xGv A, T iaov eaxo xG duo xfj? B' Xeyo, oxi eaxiv Gc; f) A Tipoc; xrjv B, ouxwc; f] B Tipoc; xf|V T. TGv yap auxGv xaxaaxeuaai!)evxwv, euel xo utio xGv A, r I'aov eaxl xG duo xfjc; B, dXXd xo drco xfjc; B xo utio xGv B, A eaxiv Tor) yap fj B xfj A- xo dpa utio xGv A, T Taov eaxl xG utio xGv B, A. eav 8e xo utio xGv axpwv i'aov fj xG utio xGv ^teawv, ai xeaaapec; eu$eTai dvdXoyov eiaiv. eaxiv dpa G? f] A Tipoc; xfjv B, ouxw<; f] A Tipoc; xf)v T. i'ar) 8e f] B xfj A- Gc; dpa f] A Tipoc; xrjv B, ouxok f] B Tipoc; xf)v T. 'Eav dpa xpelc; euiDeTai dvdXoyov Gaiv, xo utio xGv axptov Tiepiex6(ievov op-doyGviov Taov eaxl xG duo xfjc \xsor\z xe- xpayGvcp- xav xo utio xGv axpwv Tiepiexo^tevov op'doyGviov Taov fj xG aTio xfj? [Lear]z xexpayGvo, ai xpeu; eu-delai dvdXoyov eaovxai' ouep eSei 8eTc;ai. the rectangle contained by the middle (two) then the four straight-lines will be proportional. (Which is) the very thing it was required to show. Proposition 17 If three straight-lines are proportional then the rect- angle contained by the (two) outermost is equal to the square on the middle (one). And if the rectangle con- tained by the (two) outermost is equal to the square on the middle (one) then the three straight-lines will be pro- portional. A i 1 B i 1 D 1 C' 1 Let A, B and C be three proportional straight-lines, (such that) as A (is) to B, so B (is) to C. I say that the rectangle contained by A and C is equal to the square on B. Let D be made equal to B [Prop. 1.3]. And since as A is to B, so B (is) to C, and B (is) equal to D, thus as A is to B, (so) D (is) to C. And if four straight-lines are proportional then the [rectangle] contained by the (two) outermost is equal to the rectan- gle contained by the middle (two) [Prop. 6.16]. Thus, the (rectangle contained) by A and C is equal to the (rectangle contained) by B and D. But, the (rectangle contained) by B and D is the (square) on B. For B (is) equal to D. Thus, the rectangle contained by A and C is equal to the square on B. And so, let the (rectangle contained) by A and C be equal to the (square) onB. I say that as A is to B, so B (is) to C. For, with the same construction, since the (rectangle contained) by A and C is equal to the (square) on B. But, the (square) on B is the (rectangle contained) by B and D. For B (is) equal to D. The (rectangle contained) by A and C is thus equal to the (rectangle contained) by B and D. And if the (rectangle contained) by the (two) outermost is equal to the (rectangle contained) by the middle (two) then the four straight-lines are proportional [Prop. 6.16]. Thus, as A is to B, so D (is) to C. And B (is) equal to D. Thus, as A (is) to B, so B (is) to C. Thus, if three straight-lines are proportional then the rectangle contained by the (two) outermost is equal to the square on the middle (one). And if the rectangle con- tained by the (two) outermost is equal to the square on the middle (one) then the three straight-lines will be pro- portional. (Which is) the very thing it was required to 173 ETOIXEIfiN 9'. ELEMENTS BOOK 6 TCaxco f] [iev So-deTaa eui5ela f) AB, to 8e SoiSev eui9uypa[4iov to TE- 5eT Sf] duo xfj? AB eui!)e(a<; xG TE eui9uypd[4io o\ioi6v xe xal 6fio[«<; xel^ievov euiJuypaii^ov dvaypdijiai. 'ETieCeu)(T5w r) AZ, xal auveaxdxo Tip6<; xfj AB eu'dela xal xdig Ttpoc; auxfj aruieioic; xou; A, B xfj |iev upoc; xG T ycovia iar] f) Otto HAB, xfj Be uno IAZ lar\ f] hub ABH. XoiTif] dpa f) utio rZA xfj utio AHB eaxiv iar) - laoyGviov dpa eaxl xo ZrA xpiytovov xG HAB xpiyGvw. dvdXoyov dpa eaxlv (be; f) ZA Ttpog xiqv HB, ouxcoc f] Zr Tcpog xf]v HA, xal f| TA Tcpoc; xf]v AB. TidXiv auveaxdxco npog xfj BH eui5eia xal xou; Tcpoc; auxfj arpsioiz xolc; B, H xrj \ie\i utco AZE ywvia lot] f] utio BH0, xfj Se utio ZAE Tar) f] Otto HBO. Xoinf) dpa f) Ttpoc; xG E XoiTtfj xfj Ttpoc; xG eaxiv Iar)' laoyGviov dpa eaxl xo ZAE xpiyiovov xG HOB xpiyGvcy dvdXoyov dpa eaxlv Gc; f) ZA Ttp6<; xfjv HB, ouxcog f) ZE Tcpoc; xfjv HO xal f) EA Tipog xfjv OB. sSei/iSr) 8e xal Gc; f) ZA Tcpo<; xf)v HB, ouxcoc r) Zr Tcpoc; xf)v HA xal f) TA Ttpoc; xf)v AB- xal Gc; dpa f) Zr Ttpoc; xf)v AH, ouxmc; fj xe TA Ttpoc; xfjv AB xal f] ZE Ttpoc; xf)v HO xal exi f) EA Ttpoc; xf)v OB. xal eTtel Iar) eaxlv f) [xev Gito TZA y«v[a xfj utio AHB, r) 8e bub AZE xfj utio BHO, 6Xr) dpa f) utio TZE 6Xr) xfj utio AHO eaxiv Iar). 6ia xa auxd 6f) xal f] utio IAE xfj Otto ABO eaxiv iar]. eaxi 8e xal f) [iev Ttpoc; xG T xfj Ttpoc; xG A I'ar), f) 8s Ttpoc; xG E xfj Ttpoc; xG O. laoyGviov dpa eaxl xo AO xG TE- xal xac; itepl xac; laac; ycoviac; aG xGv TtXeupdc; dvdXoyov e/ei- o[ioiov dpa eaxl xo AO eu^uypajujiov xG TE eMuypd^cp. Atio xrjc; SoiSeia/jt; dpa cu^eiac; xfjg AB xG Scdevxi eu'duypdfi^cp xG TE 6y.oi6\ xe xal by.oioic, xei[ie\ov eu-duypa- y.[io\ dvayeypanxai xo AO- orcep eSei rcoifjaai. show. Proposition 18 To describe a rectilinear figure similar, and simi- larly laid down, to a given rectilinear figure on a given straight-line. Let AB be the given straight-line, and CE the given rectilinear figure. So it is required to describe a rectilinear figure similar, and similarly laid down, to the rectilinear figure CE on the straight-line AB. Let DF have been joined, and let GAB, equal to the angle at C, and ABG, equal to (angle) CDF, have been constructed on the straight-line AB at the points A and B on it (respectively) [Prop. 1.23]. Thus, the remain- ing (angle) CFD is equal to AGB [Prop. 1.32]. Thus, triangle FCD is equiangular to triangle GAB. Thus, proportionally, as FD is to GB, so FC (is) to GA, and CD to AB [Prop. 6.4]. Again, let BGH, equal to an- gle DFE, and GBH equal to (angle) FD£, have been constructed on the straight-line BG at the points G and B on it (respectively) [Prop. 1.23]. Thus, the remain- ing (angle) at E is equal to the remaining (angle) at H [Prop. 1.32]. Thus, triangle FDE is equiangular to tri- angle GHB. Thus, proportionally, as FD is to GB, so FE (is) to GB, and ED to HB [Prop. 6.4]. And it was also shown (that) as FD (is) to GB, so FC (is) to GA, and CD to AB. Thus, also, as FC (is) to ^G, so CD (is) to AB, and FE to Gi7, and, further, ED to i/B. And since angle CFD is equal to AGB, and DF£ to BCff, thus the whole (angle) CFE is equal to the whole (an- gle) AGH. So, for the same (reasons), (angle) CDE is also equal to ABH. And the (angle) at C is also equal to the (angle) at A, and the (angle) at E to the (angle) at H. Thus, (figure) AH is equiangular to CE. And (the two figures) have the sides about their equal angles pro- portional. Thus, the rectilinear figure AH is similar to the rectilinear figure CE [Def. 6.1]. Thus, the rectilinear figure AH, similar, and similarly laid down, to the given rectilinear figure CE has been constructed on the given straight-line AB. (Which is) the 174 ETOIXEIfiN 9'. ELEMENTS BOOK 6 I'd'. Ta o^ioia xpiycova npoc; dXXrjXa ev SmXaoiovi Xoycp eaxi xtbv o^oXoycov nXsupcbv. A B H T E Z 'Eaxco ojioia xpiycova xd ABr, AEZ larjv £)(ovxa xrjv npog xcp B ycoviav xfj npoc; xcp E, coc Se xrjv AB npoc; xrjv Br, ouxcoc; xrjv AE npoc; xrjv EZ, coaxe 6[i6Xoyov sTvai xrjv Br xfj EZ- Xeyco, oxi xo ABr xpiycovov npoc; xo AEZ xpiycovov 8inXaaiova Xoyov ex £1 *Fep ^1 Br npog xrjv EZ. EiXrjcp'dco yap xcov Br, EZ xpixrj dvdXoyov fj BH, coaxe elvai <b<; xrjv Br npoc; xrjv EZ, ouxcoc; xrjv EZ npoc; xrjv BH- xal STie^su/'dw rj AH. 'End ouv eaxiv cbc; rj AB npoc; xf)v Br, ouxcoc; f] AE npoc; xrjv EZ, svaXXac; apa eaxlv coc; f) AB npoc; xrjv AE, ouxcoc; f] Br npoc; xrjv EZ. dXX' cbc; rj BT npoc EZ, ouxcoc; eaxiv f) EZ npoc; BH. xal cbc; apa fj AB npoc; AE, ouxcoc; fj EZ npoc; BH- xcov ABH, AEZ apa xpiycbvcov dvxmsnovdaaiv al nXsupal ai nepi xac; loac; ycovdic;. cov Se ^iav [iia !arjv exovxeov ycoviav xpiycbvcov dvxinenovdaoiv ai nXeupal ai tie pi xac; Iaac; ycovdic;, laa eaxlv exelva. ioov apa eaxi xo ABH xpiycovov xcp AEZ xpiycbvcp. xal enei eaxiv cbc; rj Br npoc xrjv EZ, ouxcoc; rj EZ npoc; xrjv BH, eav 8e xpelc; eu$eTai dvdXoyov coaiv, fj npcoxrj npoc; xrjv xpixrjv 8mXaa[ova Xoyov e)(ei f]TC£p npoc; xrjv Seuxepav, f) Br apa npoc; xrjv BH 8mXaaiova Xoyov £X£i f^Tcep fj TB npoc; xrjv EZ. cbc; 8s f] TB npoc; xrjv BH, ouxcoc; xo ABr xpiycovov npoc; xo ABH xpiycovov xal xo ABr apa xpiycovov npoc; xo ABH 8mXaa[ova Xoyov e)(ei rjnep f\ Br npoc; xrjv EZ. Ioov 8e xo ABH xpiycovov xcp AEZ xpiycbvcp. xal xo ABr apa xpiycovov npoc; xo AEZ xpiycovov 8mXaaiova Xoyov eys\. rjnep fj Br npoc "crjv EZ. Ta apa o^toia xpiycova npoc; aXXrjXa ev 8mXaoiovi Xoyco eaxi xcov o^ioXoycov nXeupcbv. [onep eSei 8ac;ai.] liopiajjia. 'Ex 6rj xouxou cpavepov, oxi, eav xpeu; eO'delai dvdXoyov coaiv, eaxiv cbc; fj npcbxrj npoc; xrjv xpixrjv, ouxcoc; xo dno very thing it was required to do. Proposition 19 Similar triangles are to one another in the squared* ratio of (their) corresponding sides. A B G C E F Let ABC and DEF be similar triangles having the angle at B equal to the (angle) at E, and AB to BC, as DE (is) to EF, such that BC corresponds to EF. I say that triangle ABC has a squared ratio to triangle DEF with respect to (that side) BC (has) to EF. For let a third (straight-line), BG, have been taken (which is) proportional to BC and EF, so that as BC (is) to EF, so EF (is) to BG [Prop. 6.11]. And let AG have been joined. Therefore, since as AB is to BC, so DE (is) to EF, thus, alternately, as AB is to DE, so BC (is) to EF [Prop. 5.16]. But, as BC (is) to EF, so £F is to BG. And, thus, as (is) to DE, so SB (is) to BG. Thus, for triangles ABC and DEF, the sides about the equal angles are reciprocally proportional. And those triangles having one (angle) equal to one (angle) for which the sides about the equal angles are reciprocally proportional are equal [Prop. 6.15]. Thus, triangle ABG is equal to triangle DEF. And since as BC (is) to EF, so EF (is) to BG, and if three straight-lines are proportional then the first has a squared ratio to the third with respect to the second [Def. 5.9], BC thus has a squared ratio to BG with respect to (that) CB (has) to EF. And as CB (is) to BG, so triangle ABC (is) to triangle ABG [Prop. 6.1]. Thus, triangle ABC also has a squared ratio to (triangle) ABG with respect to (that side) BC (has) to EF. And triangle ABG (is) equal to triangle DEF. Thus, trian- gle ABC also has a squared ratio to triangle DEF with respect to (that side) BC (has) to EF. Thus, similar triangles are to one another in the squared ratio of (their) corresponding sides. [(Which is) the very thing it was required to show] . Corollary So it is clear, from this, that if three straight-lines are proportional, then as the first is to the third, so the figure 175 ETOIXEIfiN 9'. ELEMENTS BOOK 6 xrjc 7ipwxr](; sTSog Tipoc; xo anb xrjc; Seuxepac; xo 6[ioiov xal ojiolwc; dvaypacpo^ievov. oxerp eSei 8eic;ai. t Literally, "double". X . Td o\Loia TtoXuycova sic; xs o^oia xpiyiova 8iaipeTxai xal etc; laa xo TiXrj'doc; xal 6(i6Xoya xou; oXoic;, xal xo TtoXuycovov Tipoc; xo xoXuycovov SiTiXaaiova Xoyov *F £ P ^ o^ioXoyoc; xXeupd Tipoc; xr)v o^ioXoyov TtXsupdv. r a "Eaxw ojioia TtoXuycova xd ABrAE, ZHOKA, o^ioXoyoc; 8e eox« f] AB xrj ZH- Xsyw, oxi xd ABrAE, ZH0KA TtoXuytova eic; xe o^ioia xpiywva BiaipeTxai xal sic; 'laa xo TiXrj'doc; xal 6(i6Xoya xolc; oXoic;, xal xo ABrAE TioXuyiovov Tipoc; xo ZHOKA xoXuycovov BiTtXaaiova Xoyov e);ei rjTtep f] AB Tipoc; xr]v ZH. Tks^eux^aav ai BE, EI\ HA, A6. Kal ind o^ioiov saxi xo ABrAE TioXuycovov xfi ZHOKA TcoXuywva), Tar) eraxlv r) utio BAE yiovia xfj utco HZA. xai saxiv tbc; f] BA Tipoc; AE, ouxwc; f] HZ Tipoc; ZA. end ouv 860 xpiywvd eaxi xd ABE, ZHA (Jiav ywvlav [iia ycovia ia/]v £)(ovxa, Tispl 8e xdc; I'aac; yioviac; xdc; TiXeupdc; dvdXoyov, laoycjviov dpa eaxl xo ABE xpiycovov xfi ZHA xpiycovcp- &>ots xal o^oiov lor) dpa sraxlv r) 0x6 ABE ycovia xfj utio ZHA. saxi 8e xal oXrj f] utio ABr oXrj xrj utio ZHO lat] 8id xr]v 6[ioioxr)xa xcov TioXuytovtov Xoixr] dpa f] utio EBr y«v£a xfj utio AH© eaxiv iar\. xal etcsl 8id xrjv 6|jioL6xr)xa xwv ABE, ZHA xpiywvtov saxlv (I>c; rj EB Tipoc; BA, ouxmcj f) AH Ttpocj HZ, dXXd [ir)\ xal Sid xr|v o^ioioxrjxa xfiv TcoXuycovwv saxlv tbcj f) AB Tipoc; Br, ouxwc; f) ZH Tipoc; HO, 81' igou dpa eaxlv «c; f) EB Tipoc; Br, ouxcoc; i) AH Tipoc; H9, xal Tiepl xdc; I'oac; ycovdic; xdc; utio EBr, AH0 ai xXeupal dvdXoyov eiaiv iooycbviov dpa eaxl xo EBr xpiyovov iw AH© xpiywvo" oaxe xal 6[ioi6v iaxi xo EBr xpiycovov xfi AH© xpiytovw. 81a xd auxd 8r) xal xo ErA xpiywvov o^ioiov eoxi ifi A0K xpiywvw. xd dpa o^ioia TtoXuycova xd ABrAE, ZHOKA eic; xe o|ioia xpiycova 8if]pr)xai xal eic; laa (described) on the first (is) to the similar, and similarly described, (figure) on the second. (Which is) the very thing it was required to show. Proposition 20 Similar polygons can be divided into equal numbers of similar triangles corresponding (in proportion) to the wholes, and one polygon has to the (other) polygon a squared ratio with respect to (that) a corresponding side (has) to a corresponding side. A C D Let ABCDE and FGHKL be similar polygons, and let AB correspond to FG. I say that polygons ABCDE and FGHKL can be divided into equal numbers of simi- lar triangles corresponding (in proportion) to the wholes, and (that) polygon ABCDE has a squared ratio to poly- gon FGHKL with respect to that AB (has) to FG. Let BE, EC, GL, and LH have been joined. And since polygon ABCDE is similar to polygon FGHKL, angle BAE is equal to angle GFL, and as BA is to AE, so GF (is) to FL [Def. 6.1]. Therefore, since ABE and FGL are two triangles having one angle equal to one angle and the sides about the equal angles propor- tional, triangle ABE is thus equiangular to triangle FGL [Prop. 6.6]. Hence, (they are) also similar [Prop. 6.4, Def. 6.1]. Thus, angle ABE is equal to (angle) FGL. And the whole (angle) ABC is equal to the whole (angle) FGH, on account of the similarity of the polygons. Thus, the remaining angle EBC is equal to LGH. And since, on account of the similarity of triangles ABE and FGL, as EB is to BA, so LG (is) to GF, but also, on account of the similarity of the polygons, as AB is to BC, so FG (is) to GH, thus, via equality, as EB is to BC, so LG (is) to GH [Prop. 5.22], and the sides about the equal angles, EBC and LGH, are proportional. Thus, triangle EBC is equiangular to triangle LGH [Prop. 6.6]. Hence, triangle EBC is also similar to triangle LGH [Prop. 6.4, Def. 6.1]. So, for the same (reasons), triangle ECD is also similar 176 ETOIXEIfiN 9'. ELEMENTS BOOK 6 to 'KXfj'dog. Aeyto, oxi xal 6[i6Xoya xou; oXoic;, xouxeaxiv &axe dvdXoyov elvai xd xpiywva, xal r]yo6[jieva ^tev elvai xd ABE, EBr, ErA, eTto^ieva 8e auxwv xd ZHA, AH9, A0K, xal oxi xo ABrAE TtoXuywvov xpbc, xo ZH0KA TtoXuycovov 8iTtXaaiova Xoyov l/ei r^Ttep f) ojioXoyoc; TtXeupd Ttpoc; xr]v o^ioXoyov TtXeupdv, xouxeaxiv f] AB Ttpoc; x/]v ZH. , ETte£eu)( , dcoaav yap ai Ar, Z0. xal eitel Bid x/]v 6^oioxr]xa xwv TtoXuycovcov iar] eaxlv f] Otco ABr ycovia xfj uno ZHO, xai eaxiv £><; f] AB Ttpoc; Br, ouxwc; f) ZH Ttpoc; H9, laoytoviov eaxi xo ABr xpiywvov xw ZH0 xpiywvar tar) dpa eaxlv r] ^xev bub BAT ywvia xfj UTto HZ0, f) 8e 0ti6 BEA xfj UTto H0Z. xal CTtel larj eaxlv r] 0ti6 BAM yovia xfj UTto HZN, eaxi 8e xal f] bub ABM xfj UTto ZHN iar), xal Xomf] dpa f] UTto AMB XoiTtrj xrj UTto ZNH iar) eaxiv laoywviov dpa eaxi xo ABM xplywvov xw ZHN xpiyova). o^ioiwc; 8rj 8eT^o[iev, 6x1 xal xo BMr xpiywvov laoywviov eaxi xw HN0 xpiycovw. dvdXoyov dpa eaxiv, cbc; y.sv rj AM Ttpoc; MB, ouxcoc; f) ZN Ttpoc; NH, cbc; 8e f] BM Ttpoc; Mr, ouxoc; f) HN Ttpoc; N0- cbaxe xal 81'' I'aou, cbc; f) AM Ttpoc; Mr, ouxwc; f] ZN Ttpoc; N6. dXX' <b<; f] AM Ttpoc Mr, ouxwc; xo ABM [xpiycovov] Ttpoc; xo MBr, xal xo AME Ttpoc; xo EME Ttpoc; dXXrjXa ydp eiaiv cbc; ai paaeu;. xal cbc; dpa ev ifiv rjyou^ivwv Ttpoc; ev tuv CTtou-evcov, ouxoc; aTtavxa xd f]you^teva Ttpoc; aTtavxa xd CTtojieva' cbc; dpa xo AMB xpiywvov Ttpoc; xo BMr, ouxwc; xo ABE Ttpoc; xo TBE. aXX' cbc; xo AMB Ttpoc; xo BMr, ouxtoc; f) AM Ttpoc; Mr- xal cbc; dpa f\ AM Ttpoc; Mr, ouxcoc; xo ABE xpiywvov Ttpoc; xo EBr xplywvov. Sid xd auxd 8r] xal cbc; f) ZN Ttpoc; N0, ouxoc; xo ZHA xpiywvov Ttpoc; xo HAG xpiywvov. xai eaxiv (be; f] AM Tipoc Mr, ouxw<; f) ZN Ttpoc; N0- xal (be; dpa xo ABE xpiyovov Ttpoc; xo BEr xpiyuvov, ouxtoc; xo ZHA xpiywvov icpoc xo HA9 xpiywvov, xal evaXXd? &>c, xo ABE xpiyovov Ttp6<; xo ZHA xpiyovov, ouxoc xo BEr xpiywvov jcp6<; xo HA6 xpiywvov. ojioiclx; 8f) 8£[^o[iev eTciC£U)fdeiaa)v xwv BA, HK, oxi xal w<; xo BEr xpiywvov icpo^ xo AH0 xpiywvov, ouxw<; xo ErA xpiywvov Ttpoc; xo A0K xptywvov. xal ETtst saxtv «<; xo ABE xptywvov Ttpoc xo ZHA xptywvov, oux6X xo EBr Ttpoc; xo AH0, xal exi xo ErA Ttpoc; xo A0K, xal (be; dpa ev xwv fjyou^fvwv Ttpoc; ev xwv CTtojievcov, ouxoc aTtavxa xd fpfou\ievoL Ttpoc; aTtavxa xd STto^ieva' eaxiv dpa &>Z xo ABE xpiyovov Ttpoc; xo ZHA xpiyovov, ouxwc; xo ABrAE TtoXuywvov Ttpoc; xo ZH6KA TtoXuywvov. dXXd xo ABE xpiywvov Ttpoc; xo ZHA xpiywvov SmXaabva Xoyov eX^i ^Ttep f) AB ojioXoyoc; TtXeupd Ttpoc; xr)v ZH o^ioXoyov TtXsupdv xd yap o^ioia xpiywva ev 8iTtXaa(ovi Xoyco eaxl xwv o^ioXoywv TtXeupwv. xal xo ABrAE dpa TtoXuywvov Ttpoc; xo ZH0KA TtoXuywvov 8iTtXaaiova Xoyov e)(ei f]Ttep rj AB ojioXoyoc; TtXeupd Ttpoc; xrjv ZH o^ioXoyov TtXeupdv. Td dpa o^ioia TtoXuywva eic; xe o^ioia xptyova Biaipelxai xal el? laa xo TtXfj'doc; xal ojioXoya xoTc; oXoic;, xal xo to triangle LHK. Thus, the similar polygons ABCDE and FGHKL have been divided into equal numbers of similar triangles. I also say that (the triangles) correspond (in propor- tion) to the wholes. That is to say, the triangles are proportional: ABE, EBC, and ECD are the leading (magnitudes), and their (associated) following (magni- tudes are) FGL, LGH, and LHK (respectively) . (I) also (say) that polygon ABCDE has a squared ratio to poly- gon FGHKL with respect to (that) a corresponding side (has) to a corresponding side — that is to say, (side) AB to FG. For let AC and FH have been joined. And since angle ABC is equal to FGH, and as AB is to BC, so FG (is) to GH, on account of the similarity of the polygons, triangle ABC is equiangular to triangle FGH [Prop. 6.6]. Thus, angle BAC is equal to GFH, and (angle) BCA to GHF. And since angle BAM is equal to GFN, and (angle) ABM is also equal to FGN (see earlier), the remaining (angle) AMB is thus also equal to the remaining (angle) FNG [Prop. 1.32]. Thus, triangle ABM is equiangular to triangle FGN. So, similarly, we can show that triangle BMC is also equiangular to triangle GNH. Thus, pro- portionally, as AM is to MB, so FN (is) to NG, and as BM (is) to MC, so GN (is) to NH [Prop. 6.4]. Hence, also, via equality, as AM (is) to MC, so FN (is) to A^i? [Prop. 5.22]. But, as AM (is) to MC, so [triangle] ABM is to MBC, and AME to £MC. For they are to one an- other as their bases [Prop. 6.1]. And as one of the leading (magnitudes) is to one of the following (magnitudes), so (the sum of) all the leading (magnitudes) is to (the sum of) all the following (magnitudes) [Prop. 5.12]. Thus, as triangle AMB (is) to BMC, so (triangle) ABE (is) to CBE. But, as (triangle) AMB (is) to BMC, so AM (is) to MC. Thus, also, as AM (is) to MC, so triangle ABE (is) to triangle EBC. And so, for the same (reasons), as FN (is) to NH, so triangle FGL (is) to triangle GLH. And as AM is to MC, so i^N (is) to NH. Thus, also, as triangle ABE (is) to triangle BEC, so triangle i^GL (is) to triangle GLH, and, alternately, as triangle ABE (is) to triangle FGL, so triangle BEC (is) to triangle GLi? [Prop. 5.16]. So, similarly, we can also show, by joining BD and GK, that as triangle BEC (is) to triangle LGH, so triangle ECD (is) to triangle LHK. And since as tri- angle ABE is to triangle FGL, so (triangle) £BG (is) to LGH, and, further, (triangle) ECD to LHK, and also as one of the leading (magnitudes is) to one of the fol- lowing, so (the sum of) all the leading (magnitudes is) to (the sum of) all the following [Prop. 5.12], thus as trian- gle ABE is to triangle FGL, so polygon ABCDE (is) to polygon FGHKL. But, triangle ABE has a squared ratio 177 ETOIXEIfiN 9'. ELEMENTS BOOK 6 rcoXuyiovov rcpoc; to rcoXuywvov 8ircXaaiova Xoyov e)(ei f^TCEp f] 6[i6Xoyo<; rcXcupa rcpoc; ttjv ojioXoyov rcXcupdv [orcsp sSel 8eT^ai] . I16pia[Jia. 'Qaauxcoc; Se xal ski xfiiv [ojioiwv] xsxparcXsuptov Seix^- aexai, oxi ev BircXaafovi Xoyo elol xwv 6^oX6y«v rcXeupfiv. eBelx^t] 8e xal eni x£5v xpiycovcov wots xal xai&oXou xa ojKna eu-duypa^iia ax^liaxa rcpoc; aXXrjXa ev BircXaalovi Xoycp elol iSv ojioXoycov rcXeupwv. orcep eSei 8ei?ai. xa'. Ta iS auxfi eO'duypd^cp o^ioia xal dXXr]Xoic; eaxlv ojioia. "Eaxw yap sxdxspov xfiv A, B eO'duypd^cov iw T o^ioiov Xsy«, oxi xal xo A tw B eaxiv o^oiov. 'Ercri yap o^ioiov eaxi xo A xa> T, laoycoviov xe eaxiv auxw xal xac; rcepl xac; laag ycoviac; rcXeupac; dvdXoyov ey^ei. rcdXiv, end ojioiov eaxi xo B x£5 T, laoycoviov xe eaxiv auxo xal xac; rcepl xac; Taac; ycoviac; rcXeupac; dvdXoyov ex £l - exdxepov apa xwv A, B xG T icoywviov xe eaxi xal xac; rcepl xac; taac; ywviac; rcXeupac; dvdXoyov s^ei [waxe xal xo A tu B laoycoviov xe eaxi xal xac; rcspl xac foat; ytoviac; to triangle FGL with respect to (that) the corresponding side AB (has) to the corresponding side FG. For, similar triangles are in the squared ratio of corresponding sides [Prop. 6.14]. Thus, polygon ABCDE also has a squared ratio to polygon FGHKL with respect to (that) the cor- responding side AB (has) to the corresponding side FG. Thus, similar polygons can be divided into equal num- bers of similar triangles corresponding (in proportion) to the wholes, and one polygon has to the (other) polygon a squared ratio with respect to (that) a corresponding side (has) to a corresponding side. [(Which is) the very thing it was required to show] . Corollary And, in the same manner, it can also be shown for [similar] quadrilaterals that they are in the squared ratio of (their) corresponding sides. And it was also shown for triangles. Hence, in general, similar rectilinear figures are also to one another in the squared ratio of (their) corre- sponding sides. (Which is) the very thing it was required to show. Proposition 21 (Rectilinear figures) similar to the same rectilinear fig- ure are also similar to one another. Let each of the rectilinear figures A and B be similar to (the rectilinear figure) C. I say that A is also similar to B. For since A is similar to C, {A) is equiangular to (C), and has the sides about the equal angles proportional [Def. 6.1]. Again, since B is similar to C, (B) is equian- gular to (C), and has the sides about the equal angles proportional [Def. 6.1]. Thus, A and B are each equian- gular to C, and have the sides about the equal angles 178 ETOIXEIfiN 9'. ELEMENTS BOOK 6 TtXeupdc; dvdXoyov exei]. o^ioiov dpa sail to A tw B- oTtep e8e:i 8eTc;ai. x(3'. 'Edv xeaaapec; eu^eTai dvdXoyov SSmv, xal xd an auxfiv eGTJuypajji^a opioid xe xdi ojioicoc; dvayeypa^eva dvdXoyov eaxai- xdv xd dm' auxwv et/duypaii^ia opioid xe xal o^oico? dvayeypa^eva dvaXoyov rj, xdi auxdi ad £0156101 dvdXoyov eaovxai. A B r A n p "Eaxwaav xeaaapec; euiMai dvdXoYov ai AB, TA, EZ, H0, (be; f) AB Ttpoc; xr)v TA, ouxgk r| EZ Ttpoc; xrjv H6, xdi dvayeypdcpiSioaav a7t0 ^ v T " v AB, TA opioid xe xdi ojioiioc; xei^xeva eMuypajjijia xd KAB, ArA, aTto Se xov EZ, H9 opioid xe xdi o^ioicoc; xei^ieva eMuypa^a xd MZ, N6- Xeyw, oxi eaxiv foe, xo KAB Ttpoc; xo ArA, ouxgjc; xo MZ Ttpoc; xo N6. EiXricp-dw yap x«v |iev AB, TA xpixr) dvdXoyov f] S, xfiv 8e EZ, H9 xpixr) dvdXoyov f] 0. xdi enei eaxiv tbc; [iev f) AB Ttpoc; x/)v TA, ouxioc; f) EZ Ttpoc; xrjv H6, tbc; 8e f) TA Ttpoc; xr]v S, oux6K f] H6 Ttpoc; x/)v 0, 8i' laou dpa eaxiv (be; f) AB Ttpoc; xr)v S, ouxwg f) EZ Ttpoc; xrjv 0. dXX' tbc; jiev f] AB Ttpoc; x/)v S, ouxtoc; [xal] xo KAB Ttpoc; xo ArA, cbc; 8e f] EZ Ttpoc; x/)v 0, ouxwc; xo MZ Ttpoc; xo N6- xal (be; dpa xo KAB Ttpoc; xo ArA, ouxwc; xo MZ Ttpoc; xo N0. AXXd Sr) eaxto tbc; xo KAB Ttpoc; xo ArA, ouxwc; xo MZ Ttpoc; xo N9- Xeyw, oxi eaxl xal foe, rj AB Ttpoc; xrjv TA, ouxwc; f) EZ Ttpoc; xrjv H9. ei yap \ir\ eaxiv, foe, f) AB Ttpoc; xrjv TA, ouxwc; r) EZ Ttpoc; xrjv H9, eoxw (be; f] AB Ttpoc; xrjv TA, ouxoc r| EZ Ttpoc; xrjv IIP, xal dvayeypdcpiJo dito xrjc; proportional [hence, A is also equiangular to B, and has the sides about the equal angles proportional]. Thus, A is similar to B [Def. 6.1] . (Which is) the very thing it was required to show. Proposition 22 If four straight-lines are proportional then similar, and similarly described, rectilinear figures (drawn) on them will also be proportional. And if similar, and similarly described, rectilinear figures (drawn) on them are pro- portional then the straight-lines themselves will also be proportional. A BCD Q R Let AB, CD, EF, and GH be four proportional straight-lines, (such that) as AB (is) to CD, so EF (is) to GH. And let the similar, and similarly laid out, rec- tilinear figures KAB and LCD have been described on AB and CD (respectively), and the similar, and similarly laid out, rectilinear figures MF and NH on EF and GH (respectively). I say that as KAB is to LCD, so MF (is) to NH. For let a third (straight-line) O have been taken (which is) proportional to AB and CD, and a third (straight-line) P proportional to EF and GH [Prop. 6.11]. And since as AB is to CD, so EF (is) to GH, and as CD (is) to 0, so GH (is) to P, thus, via equality, as AB is to 0, so EF (is) to P [Prop. 5.22]. But, as AB (is) to 0, so [also] KAB (is) to LCD, and as EF (is) to P, so MF (is) to NH [Prop. 5.19 corr.]. And, thus, as KAB (is) to LCD, so MF (is) to NH. And so let KAB be to LCD, as MF (is) to NH. I say also that as AB is to CD, so EF (is) to Gi7. For if as AB is to CD, so £F (is) not to GH, let ylB be to CD, as £F 179 ETOIXEIfiN 9'. ELEMENTS BOOK 6 IIP oiioxepcp xfiv MZ, N9 o^oiov xe xal o^ioicoc; xei|jievov eu'duypa^tuov to EP. 'Ercel ouv eaxiv (be; f] AB 7ipo<; xrjv TA, ouxw<; f] EZ upoc; x/]v IIP, xal dvayeypaTixai a7l ° ^ v T " v AB, TA opioid xe xal 6(ioia>c xeifieva xd KAB, ABA, duo 8e xebv EZ, nP Sfioid xe xal o^otox xei^ieva xd MZ, SP, eaxiv apa cbc; xo KAB Tip6<; xo ArA, ouxclx; xo MZ npbc, xo SP. tmoxeixai 8e xal <b<; xo KAB 7ipo<; xo ArA, ouxwc; xo MZ 7ipo<; xo N6- xal cbc; apa xo MZ Ttpoc. xo EP, ouxwe. xo MZ npbc, xo N0. xo MZ apa Ttpoc, exdxepov xwv N0, SP xov auxov §XCi Xoyov laov apa eaxl xo N0 tw SP. eaxi 8s auxw xal o^toiov xal b[io'ux>z xei^tevov I'ar) apa f] H9 xrj nP. xal ernei eaxiv cb<; f] AB Ttpoc. xrjv TA, ouxck f) EZ Ttpoc. xrjv nP, iar] 8s f] nP xrj H0, eaxiv apa (be. f) AB Tcpoc. xrjv PA, ouxoc. rj EZ Ttpoc, xr]v H9. 'Edv apa xeaaapec, eu'delai dvdXoyov waiv, xal xd dm' auxCSv sO'duypaji^.a opioid xe xal 6^oio<; dvayeypa^eva dvdXoyov eaxar xav xd dm' auxebv eCcvMypamia opioid xe xal 6\ioi(x>q, dvayeypa^eva dvdXoYov fj, xal auxai al eu-fMai dvdXoYov eaovxai' onep eSei 8eTc;ai. (is) to QR [Prop. 6.12]. And let the rectilinear figure SR, similar, and similarly laid down, to either of MF or NH, have been described on QR [Props. 6.18, 6.21]. Therefore, since as AB is to CD, so EF (is) to QR, and the similar, and similarly laid out, (rectilinear fig- ures) KAB and LCD have been described on AB and CD (respectively), and the similar, and similarly laid out, (rectilinear figures) MF and SR on EF and QR (re- sespectively), thus as KAB is to LCD, so MF (is) to SR (see above). And it was also assumed that as KAB (is) to LCD, so MF (is) to NH. Thus, also, as MF (is) to SR, so MF (is) to NH [Prop. 5.11]. Thus, MF has the same ratio to each of NH and SR. Thus, AT? is equal to SR [Prop. 5.9]. And it is also similar, and similarly laid out, to it. Thus, GH (is) equal to QRJ And since AB is to CD, as EF (is) to QR, and QR (is) equal to GH, thus as AB is to CD, so EF (is) to Cff. Thus, if four straight-lines are proportional, then sim- ilar, and similarly described, rectilinear figures (drawn) on them will also be proportional. And if similar, and similarly described, rectilinear figures (drawn) on them are proportional then the straight-lines themselves will also be proportional. (Which is) the very thing it was required to show. t Here, Euclid assumes, without proof, that if two similar figures are equal then any pair of corresponding sides is also equal. xy'. Td iaoy(bvia TiapaXXr]X6Ypajj.(jia npbc; dXXrjXa Xoyov e)(ei xov auyxei^ievov ex xGv TiXeupcbv. 'Eaxto iaoy(bvia TtapaXXr)X6ypa|ji[j.a xd Ar, TZ Tarjv E^ovxa xrjv utio BTA ywviav xrj utio EITT Xeycj, oxi xo Ar TiapaXXr)X6ypajj.|jiov Ttpoc, xo TZ TtapaXXr)X6ypa(i|j.ov Xoyov e/ei xov auyxeijievov ex x(5v TtXeupfiiv. Kslcrdco yap (baxe in cu^eiac, elvai xr)v Br xrj TH- in eMeiac. apa eaxl xal f] AT xrj TE. xal au[iTteTtXY]p(baTL>cd xo AH 7iapaXXr]X6ypa[ji[j.ov, xal exxeicrdGj xic, cu-dela f] K, xal yeyovexw (be. \xe\ f] BT Ttpoc, xy]v TH, ouxmc. f) K npbc, xrjv A, (be 8s f] AT npbc, x/)v TE, ouxioc; f) A npog x/)v M. Oi apa Xoyoi xfjg xs K upoc; xrjv A xal xrjc; A upoc xr)v M oi auxoi eiai zoic, Xoyoic; x«v uXsupcbv, xfjc ^£ BT Tipog xrjv TH xal xrjg Ar Ttpoc; xr)v TE. dXX' 6 xfjc; K Tipoc M Xbyoc, auyxeixai ex xe xou jy]z K up6(; A Xoyou xal xou xfj<; A 7ip6<; M- tbaxe xal f) K 7tpo<; xf]v M Xoyov exei xov auyxel^evov ex xebv nXeupfiv. xal enei eaxiv cbi; f) Br Tipog xr]v TH, ouxtog xo Ar TiapaXXrjXoypa^^tov Tip6(; xo TQ, dXX' <bg f) BT rcpoc; xf)v TH, ouxoc; f) K upog xrjv A, xal (b<; apa f) K Ttpoc; xt]v A, oux«c xo AT npbz xo T6. TidXiv, eitei eaxiv ok f\ AT npbz xr]v TE, ouxwc; xo TQ ita- paXX/]X6ypa^i[iov npbc, xo TZ, dXX' cbg f] Ar 7tp6<; x/]v TE, Proposition 23 Equiangular parallelograms have to one another the ratio compounded^ out of (the ratios of) their sides. Let AC and CF be equiangular parallelograms having angle BCD equal to ECG. I say that parallelogram AC has to parallelogram CF the ratio compounded out of (the ratios of) their sides. For let BC be laid down so as to be straight-on to CG. Thus, DC is also straight-on to CE [Prop. 1.14]. And let the parallelogram DG have been completed. And let some straight-line K have been laid down. And let it be contrived that as BC (is) to CG, so K (is) to L, and as DC (is) to CE, so L (is) to M [Prop. 6.12]. Thus, the ratios of K to L and of L to M are the same as the ratios of the sides, (namely), BC to CG and DC to CE (respectively). But, the ratio of K to M is com- pounded out of the ratio of K to L and (the ratio) of L to M. Hence, K also has to M the ratio compounded out of (the ratios of) the sides (of the parallelograms). And since as BC is to CG, so parallelogram AC (is) to GH [Prop. 6.1], but as BC (is) to CG, so K (is) to L, thus, also, as K (is) to L, so (parallelogram) AC (is) to GH. Again, since as DC (is) to CE, so parallelogram 180 ETOIXEIfiN 9'. ELEMENTS BOOK 6 outwc; f] A 7ip6<; x/]v M, xdx &>c, dpa f] A npbc, ttjv M, outox to T9 TtapaXX/jXoYpa^ov Tcpoc to TZ 7tapaXXr)X6Ypa^ov. etieI ouv eBeix'dr), cbc; uev f] K npog ttjv A, outck to Ar 7iocpaXXr]X6Ypoc^ov 7ip6<; to TO 7tapaXX/]XoYpa^ov, w<; Be: f) A TCpog t/)v M, ouTtx; to T0 7iapaXX/]XoYpa|ji[j.ov upoc; to TZ 7tapaXXr]X6Ypa^[iov, 81' Taou apa eotIv cbc; f) K Ttpoc; tt]v M, outmc; to Ar Ttpoc; to TZ TtapaXXrjXoYpa^ov. f] 8e K 7ip6<; Tf]v M Xoyov g)(ei tov auyxdjisvov ex t£>v uXsupwv xal to Ar apa icpoc; to TZ Xbyo\ s/ei tov a\jyxei[ievov tx iwv uXeupOv. CF (is) to CF [Prop. 6.1], but as DC (is) to CE, so L (is) to M, thus, also, as £ (is) to M, so parallelogram CH (is) to parallelogram CF. Therefore, since it was shown that as K (is) to L, so parallelogram AC (is) to parallelogram CH, and as L (is) to M, so parallelogram CH (is) to parallelogram CF, thus, via equality, as K is to M, so (parallelogram) AC (is) to parallelogram CF [Prop. 5.22]. And K has to M the ratio compounded out of (the ratios of) the sides (of the parallelograms) . Thus, (parallelogram) AC also has to (parallelogram) CF the ratio compounded out of (the ratio of) their sides. D H K' ' A' ' Mi ' j E Z Ta apa laoYiovia TcapaXX/]XoYpa[i^.a Tipoc; aXX/]Xa Xoyov £/si tov auYxeijievov ex x&m uXeupfiv ouep e8ei 8eT<;ai. B K ' 1 L ' ' Mi 1 E F Thus, equiangular parallelograms have to one another the ratio compounded out of (the ratio of) their sides. (Which is) the very thing it was required to show. t In modern terminology, if two ratios are "compounded" then they are multiplied together. x5'. TlavToc; 7iapaXX/]XoYpa[i^ou Ta itep! xr)v Sid^ieTpov ua- paXXrjXoYpa^a opioid taxi x& ts 6X« xal dXXr|Xou;. "EaTW TiapaXXrjXoYpa^i^ov to ABTA, 8id^£Tpoc; 8e auToO f) AT, Ttepl 8s ttjv AT TcapaXXrjXoYpa^a eaTW Ta EH, 6K- Xeyw, oti cxaTspov t£Sv EH, 0K TtapaXXr)XoYpd^«v 5\ioi6v taxi SXfc) to ABTA xal dXXfiXou;. 'Etc! yap TpiY^vou tou ABT Ttapa (jiiav xSv icXeupwv t/]v BT rjxTai f] EZ, dvdXoYov icmv cbc; f] BE icpoc; t/]v EA, outca; f) TZ Ttpoc" ttjv ZA. icdXiv, etcI TpiYtbvou tou ATA icapa ^i(av ttjv TA rjxTai f\ ZH, dvdXoYov sgtiv ok f\ TZ npbc, tt]v ZA, outcx f] AH Ttpoc" ttjv HA. dXX'' cbc; r) TZ Ttpoc; t/]v ZA, outck eBdx'dr) xal r) BE Ttpoc; t/]v EA- xal cbc; apa f] BE Ttpoc; xf\v EA, outcoc; f] AH Tipoc; xf\v HA, xal auvdfvTi apa cbc; f] BA Ttpoc; AE, outcx r) AA Ttpoc; AH, xal evaXXac; cbc; r) BA Ttpoc; ttjv AA, outcoc; f) EA Ttpoc; tt]v AH. tc5v apa ABTA, EH TtapaXXr)XoYpd^«v dvdXoYov eiaiv ai TtXeupal al Ttepl Trjv xoiv/]v Y^viav ttjv utco BAA. xal sttei 7iapdXXr)X6(; ecrciv 1] HZ Tfj AT, lay] eotIv f] ^iev U7i6 AZH yav'ia xt] bub ATA- xal xoivr) xuv 8uo Proposition 24 In any parallelogram the parallelograms about the di- agonal are similar to the whole, and to one another. Let ABCD be a parallelogram, and AC its diagonal. And let EG and HK be parallelograms about AC. I say that the parallelograms EG and HK are each similar to the whole (parallelogram) ABCD, and to one another. For since EF has been drawn parallel to one of the sides BC of triangle ABC, proportionally, as BE is to EA, so CF (is) to FA [Prop. 6.2]. Again, since FG has been drawn parallel to one (of the sides) CD of trian- gle ACD, proportionally, as CF is to FA, so DG (is) to GA [Prop. 6.2]. But, as CF (is) to FA, so it was also shown (is) BE to EA. And thus as f?F (is) to EA, so DG (is) to GA. And, thus, compounding, as BA (is) to AE, so DA (is) to AG [Prop. 5.18]. And, alternately, as BA (is) to AD, so EA (is) to AG [Prop. 5.16]. Thus, in parallelograms ABCD and EG the sides about the common angle BAD are proportional. And since GF is parallel to DC, angle AFC is equal to DGA [Prop. 1.29]. 181 ETOIXEIfiN 9'. ELEMENTS BOOK 6 xpiyovwv xfiv AAr, AHZ f] uko A AT ywvla' laoywviov apa taxi xo AAr xpiycovov xfi AHZ xpiywvw. 8id xa auxa 8r) xal xo ArB xpiywvov laoycoviov saxi iu AZE xpiyovw, xal oXov xo ABrA 7iapaXXr]X6ypa^ov iu EH TtapaXXr]- Xoypd^up laoycoviov saxiv. dvdXoyov apa laxlv cot; f) AA 7ip6<; xrjv Ar, oux«<; f] AH 7tp6<; xr)v HZ, to? Ss f] Ar npoz xr]v TA, oOxto? f] HZ 7tp6<; xrjv ZA, d>c 8s f] Ar upot; x/]v TB, ouxw<; f] AZ Ttp6<; xrjv ZE, xal sxi cb<; f] TB 7tp6<; xrjv BA, oux«<; f] ZE Tipoc; xrjv EA. xal sitsl eSsix^ t^v f] Ar Ttpoc; xrjv TA, ouxw<; r\ HZ Ttpoc; xrjv ZA, £><; 8s f) Ar npd<z ir\\> TB, ouxwg f] AZ 7tp6<; xrjv ZE, 8i' laou apa saxlv cb<; r) Ar 7ip6<; xrjv TB, ouxtoc; f) HZ Ttp6<; xrjv ZE. xfiiv apa ABrA, EH 7iapaXXr)Xoypd^.[i«v dvaXoyov siaiv ai uXsupal ai nspi xdc; laac ycoviac ojioiov apa saxl xo ABrA 7iapaXXr]Xoypd^ov xw EH 7tapaXXr]Xoypd^tp. 8id xd auxa 8r) xo ABrA 7tapaXXr]X6ypa(i[jiov xal xw KO rca- paXXr)Xoypd^i[iw ojioiov saxiv sxdxspov apa xwv EH, 8K 7tapaXXr]Xoypd^tov iu ABrA [n:apaXXr)Xoypd^w] o^ioiov saxiv. xa 8s xw auxw suTJuypd^tp o^ioia xal dXXrjXoic; saxlv ofjioia' xal xo EH apa TtapaXXr)X6ypa^ov x£> 0K TtapaXXrj- Xoypd[i^tw o\±oio\i saxiv. A E B A K r navxog apa TtapaXXr)Xoypd^ou xd uspl xiqv 8id^isxpov TtapaXXiqXoypa^a ojioid saxi xai xs oXo xal dXXrjXoic oTtsp s8si SsT^ai. xs'. Tcp SotJevxi su$uypd[i[jiw 6|ioiov xal aXXtp iw So-dsvxi I'aov xo auxo auaxr]aaa , dai. And angle DAC (is) common to the two triangles ADC and AGF. Thus, triangle ADC is equiangular to triangle AGF [Prop. 1.32]. So, for the same (reasons), triangle ACB is equiangular to triangle AFE, and the whole par- allelogram ABCD is equiangular to parallelogram EG. Thus, proportionally, as AD (is) to DC, so AG (is) to GF, and as DC (is) to CA, so GF (is) to FA, and as AC (is) to CB, so AF (is) to FF, and, further, as CB (is) to BA, so F F (is) to EA [Prop. 6.4]. And since it was shown that as DC is to CA, so GF (is) to FA, and as AC (is) to CB, so (is) to FE, thus, via equality, as DC is to CF, so GF (is) to FF [Prop. 5.22]. Thus, in parallelograms ABCD and EG the sides about the equal angles are proportional. Thus, parallelogram ABCD is similar to parallelogram EG [Def. 6.1]. So, for the same (reasons), parallelogram ABCD is also similar to par- allelogram KH. Thus, parallelograms EG and HK are each similar to [parallelogram] ABCD. And (rectilin- ear figures) similar to the same rectilinear figure are also similar to one another [Prop. 6.21]. Thus, parallelogram EG is also similar to parallelogram HK. A E B D K C Thus, in any parallelogram the parallelograms about the diagonal are similar to the whole, and to one another. (Which is) the very thing it was required to show. Proposition 25 To construct a single (rectilinear figure) similar to a given rectilinear figure, and equal to a different given rec- tilinear figure. 182 ETOIXEIfiN 9'. ELEMENTS BOOK 6 A E M 'Egxm to ^t£v BoiJev £ui56ypa|ji[iov, S 8a ojiolov auaxrpacrdai, to ABr, G 8e 8eT I'aov, to A- Sa 8rj iu [lev ABr o^ioiov, iS 8e A laov to auxo ouaxr]oaa , dai. napapepXf]a , dw yap Ttapd [Lev xr)v Br xG ABr xpiycovw laov 7iapaXXr]X6Ypa^ov xo BE, Ttapd 8e xrjv TE iS A '(aov napaXXrjXoYpa^ov xo TM sv yojvia xfj Otto ZrE, fj eaxiv Tar) xfj uiio TBA. in' ed-Qsiac, dpa saxlv f) ^ev Br xfj TZ, f| 8e AE x^ EM. xal eiXfjcp'dw xfiv Br, TZ y.eor) dvaXoyov f) H9, xal dvaysypdcp^Gj arco xfjc H9 xfi ABr ojiolov xe xal o^lolmc xdjisvov xo KH9. Kal STiei saxiv foe, i] BT npbc, xfjv H9, oux«<; r) H9 upog xr]v rZ, sdv 8e xpelc; sMelai dvaXoyov fiaiv, eaxiv foe, f] upwxr] Tipoc; xf]v xpixrjv, ouxclx; xo arco xfj? Ttpt5xr]<; eT8o<; Ttpoc; xo duo xfjc 8euxepa<; xo 6[ioiov xal o^oiox dva- Ypacpouevov, eaxiv dpa ox rj Br Ttp6<; xf]v TZ, ouxgx xo ABr xpiywvov Ttp<x xo KH9 xp(y«vov. dXXd xal ox f] BT 7ip6<; xf|V rZ, ouxox xo BE TiapaXX^Xoypa^ov Ttpcx xo EZ 7iapaXXr]X6Ypa[i^ov. xal ox apa xo ABr xpiycovov Ttpoc xo KH9 xpiywvov, ouxox xo BE TtapaXXrjXoYpa^ov Ttpcx xo EZ TtapaXX/]XoYpa^ov evaXXdc; apa ox xo ABr xpiywvov Ttp<x xo BE TiapaXX^Xoypa^^ov, ouxox xo KH9 xpiYO)vov 7ip6<; xo EZ 7iapaXXr]X6Ypa^ov. taov 8e xo ABr xpiYOJvov xo BE TtapaXX/]XoYpd^oy laov dpa xal xo KH9 xpiYWvov xfi> EZ n;apaXXr)XoYpd^[io. dXXd xo EZ TiapaXX/jXoYpa^ov tu A eaxiv Taov xal xo KH9 apa xw A eaxiv iaov. eaxi 8e xo KH9 xal x« ABr ojioiov. Top apa BoiJevxi eui!)uYpd|i^oj xoj ABr opioiov xal aXXo) iG BoiJevxi iS A I'aov xo auxo auveaxaxai xo KH9- oTtep e8ei Ttoirjaai. Xf'. Eav duo TiapaXXrjXoYpd^ou TiapaXX/jXoYpa^iov dcpai- pedf 6[ioi6v xe x« 0X0) xal opioiox xei^ievov xoivfjv Y^viav eXov auxo), Ttepl xf|v auxfjv 8id^exp6v eaxi xo3 oXo>. Ako yap KapaXXrjXoYpd^^iou xou ABrA TtapaXXrjXoYpa- (iuov dcp/jprjaOw xo AZ ojioiov iu ABrA xal 6^io(w<; xsi^ievov xoivrjv Y«viav £)(ov auxw xrjv utio AAB- Xey", K L E M Let ABC be the given rectilinear figure to which it is required to construct a similar (rectilinear figure), and D the (rectilinear figure) to which (the constructed figure) is required (to be) equal. So it is required to construct a single (rectilinear figure) similar to ABC, and equal to D. For let the parallelogram BE, equal to triangle ABC, have been applied to (the straight-line) BC [Prop. 1.44], and the parallelogram CM, equal to D, (have been ap- plied) to (the straight-line) CE, in the angle FCE, which is equal to CBL [Prop. 1.45]. Thus, BC is straight-on to CF, and LE to EM [Prop. 1.14]. And let the mean pro- portion GH have been taken of BC and CF [Prop. 6.13] . And let KGB, similar, and similarly laid out, to ABC have been described on GH [Prop. 6.18]. And since as BC is to GH, so GH (is) to CF, and if three straight-lines are proportional then as the first is to the third, so the figure (described) on the first (is) to the similar, and similarly described, (figure) on the second [Prop. 6.19 corn], thus as BC is to CF, so triangle ABC (is) to triangle KGH. But, also, as BC (is) to CF, so parallelogram BE (is) to parallelogram EF [Prop. 6.1]. And, thus, as triangle ABC (is) to triangle KGH, so par- allelogram BE (is) to parallelogram EF. Thus, alter- nately, as triangle ABC (is) to parallelogram BE, so tri- angle KGH (is) to parallelogram EF [Prop. 5.16]. And triangle ABC (is) equal to parallelogram BE. Thus, tri- angle KGH (is) also equal to parallelogram EF. But, parallelogram EF is equal to D. Thus, KGH is also equal to D. And KGH is also similar to ABC. Thus, a single (rectilinear figure) KGH has been con- structed (which is) similar to the given rectilinear figure ABC, and equal to a different given (rectilinear figure) D. (Which is) the very thing it was required to do. Proposition 26 If from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the sub- tracted parallelogram) is about the same diagonal as the whole. For, from parallelogram ABCD, let (parallelogram) 183 ETOIXEIfiN 9'. ELEMENTS BOOK 6 6xi uepl xrjv auxrjv Sidjiexpov eaxi to ABrA tw AZ. AH A b r Mr) yap, dXX' el 8uvax6v, eax« [auxfiv] 8id^iexpo<; fj AQr, xal expXrrdeTaa fj HZ SirjX'dw etxi xo 9, xal rj)cdco Sid xou oitopepa iwv AA, Br TtapdXXrjXoc; fj 0K. 'Excel ouv uepl xrjv auxrjv Sidjuexpov eaxi xo ABrA iw KH, eaxiv dpa (be; rj A A npbz xrjv AB, ouxioc; f) HA upog xrjv AK. eaxi 8e xal 8ia xrjv o^toioxrjxa xwv ABrA, EH xal cb<; f) AA 7ip6<; xrjv AB, ouxck f] HA Tcp6<; xrjv AE- xal &>c, dpa f) HA npbc, xrjv AK, ouxcoc; fj HA npoc; xrjv AE. fj HA dpa TCpog exaxepav xfiv AK, AE xov auxov e/ei Xoyov. larj dpa eaxlv fj AE xfj AK fj eXdxxov xfj ^eii^ovi- orcep laxlv dBuvaxov. oux dpa oux eaxi uepl xrjv auxrjv 8id[uexpov xo ABrA xw AZ- uepl xrjv auxrjv dpa eaxi 8id^expov xo ABrA TiapaXXrjXoypa^ov xw AZ TtapaXXrjXoYpd^cp. 'Edv dpa diio 7tapaXXrjXoYpd^[Uou TiapaXXrjXoypa^iuov dcpaipcdfj 6|ioi6v xe x« 6Xo xal b[io'ux>z xeijievov xoivrjv yoviav s^ov auxG, rcepl xrjv auxrjv Siduexpov eoxi x£3 6Xcp- oTiep e8ei 8eT^ai. ndvxwv xwv iiapd xrjv auxrjv eMeTav napapaXXo^ievwv TiapaXXrjXoYpd^i[Uwv xal eXXeinovxov eiSeai KapaXXrjXoypdfU- [ioic by.oioic, xe xal 6iuoio<; xeiiuevou; xw aito xfj? fjiuiaeiac; dvaypacpotuevw [ueyiaxov eaxi xo duo xff. rj^iiaeiac; napa- PaXXo^tevov [TtapaXXrjXoYpa^ov] o^ioiov ov iu eXXei^ifiavxi. "Eaxw eu'deTa fj AB xal xexurja'do "biya. xaxa xo T, xal TiapapepXrjadw Ttapa xrjv AB eu'delav xo AA uapaX- XrjXoYpaji^ov eXXelnov ei8ei TCapaXXrjXoypd^o xw AB dva- Ypacpevxi duo xrj<; rj[iiae(a<; xfj<; AB, xouxeaxi xrj<; TB - Xeya>, oxi udvxwv xwv Ttapa xrjv AB uapapaXXo^evwv TtapaXXrj- XoYpd^[Uwv xal eXXeiKovxwv eiSeai [TtapaXXrjXoypdfU^ou;] oiuoiok; xe xal 6uo(m<; xeiuevoic; xo AB ^icyiaxov eaxi xo AF have been subtracted (which is) similar, and similarly- laid out, to ABCD, having the common angle DAB with it. I say that ABCD is about the same diagonal as AF. AG D B C For (if) not, then, if possible, let ARC be [ABCD's] diagonal. And producing GF, let it have been drawn through to (point) H. And let HK have been drawn through (point) H, parallel to either of AD or BC [Prop. 1.31]. Therefore, since ABCD is about the same diagonal as KG, thus as DA is to AB, so GA (is) to AK [Prop. 6.24]. And, on account of the similarity of ABCD and EG, also, as DA (is) to AB, so GA (is) to AE. Thus, also, as GA (is) to AK, so GA (is) to AE. Thus, GA has the same ratio to each of AK and AE. Thus, AE is equal to AK [Prop. 5.9], the lesser to the greater. The very thing is impossible. Thus, ABCD is not not about the same di- agonal as AF. Thus, parallelogram ABCD is about the same diagonal as parallelogram AF. Thus, if from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the subtracted parallelogram) is about the same diagonal as the whole. (Which is) the very thing it was required to show. Proposition 27 Of all the parallelograms applied to the same straight- line, and falling short by parallelogrammic figures similar, and similarly laid out, to the (parallelogram) described on half (the straight-line), the greatest is the [parallelo- gram] applied to half (the straight-line) which (is) similar to (that parallelogram) by which it falls short. Let AB be a straight-line, and let it have been cut in half at (point) C [Prop. 1.10]. And let the parallelogram AD have been applied to the straight-line AB, falling short by the parallelogrammic figure DB (which is) ap- plied to half of AB — that is to say, CB. I say that of all the parallelograms applied to AB, and falling short by 184 ETOIXEIfiN 9'. ELEMENTS BOOK 6 AA. Tiocpotpepxyja'dco yap Tiapd xrjv AB euifteTav to AZ na- paXX/]X6ypa^ov eXXeinov d'Bsi TiapaXX/jXoypd^cp to ZB o^oto xe xal o^oiwc xei^tevw xcp AB- Xeyto, oxi uei£6v eaxi xo AA xou AZ. A E A r K B 'Etc! yap ojioiov eaxi xo AB TtapaXXrjXoypa^ov xG ZB 7iapaXX/]Xoypd[i^.CL), Ttepl x/]v auxr|v eiai 8id[iexpov. f))fdw auxwv 8id^.expo<; f] AB, xal xaxayeypdcp-dw xo ax?j^a. 'Ercel o5v iaov eaxi xo TZ xw ZE, xoivov Be xo ZB, oXov dpa xo T6 oX« xw KE eaxiv iaov. dXXd xo T6 tu TH eaxiv iaov, etiei xal f] Ar xfj TB. xal xo BT dpa x£S EK eaxiv iaov. xoivov Ttpoaxe(a , da> xo TZ' oXov dpa xo AZ iu AMN yvw^tovi eaxiv iaov waxe xo AB TtapaXXr)X6ypa^iov, xouxeaxi xo AA, xou AZ 7iapaXX/]Xoypd[i^ou \±s%6v eaxiv. ndvxcov dpa xwv Ttapa xrjv auxrjv eu'delav TtapapaX- Xo^tevwv TiapaXXr)Xoypd^«v xal eXXeiTtovxov eiBeai TtapaX- Xr]Xoypd^oi<; b\±oioiq xe xal 6\±oig>z xei^iivou; iw duo xfj<; f|^iaeia<; dvaypacpo^tevo ^teyiaxov eaxi xo duo xrj<; f^iaeiac; 7iapapX/]Tf)£v orcep e8ei BeT^ai. XT]'. Ilapd xiqv 8oi9eTaav euiMav xw Bo'devxi eu'duypd^w Iaov napaXXiqXoypa^ov napapaXelv eXXeutov ei8ei na- paXX/)Xoypd|jijicp 6|ioicp xw BoiSevxi- Bel Be xo BiBopievov eO'duypa^ov [S BeT iaov TtapapaXeTv] ^ir] ^xeT^ov eivai xou duo xrjg rpiaeiag dvaypacpo^ievou o^ioiou x« eXXei^axi [xou xe duo xfj? f)|jiiaeia<; xal 5 Bel o^ioiov eXXeineiv] . 'Eaxco f) (iev So-deTaa eui)eTa f) AB, xo Be Bo-dev eu-duypajijiov, £> BeTTaov Ttapd xrjv AB TiapapaXelv, xo T [if] ^el^ov [dv] xou dno xfjc; fj^iiaeiac; xfjg AB dvaypacpo^ievou o^tobu ifi eXXei^axi, S Be: BeT o^ioiov eXXeiTteiv, xo A- 8a Br) [parallelogrammic] figures similar, and similarly laid out, to DB, the greatest is AD. For let the parallelogram AF have been applied to the straight-line AB, falling short by the parallelogrammic figure FB (which is) similar, and similarly laid out, to DB. I say that AD is greater than AF. D E A C K B For since parallelogram DB is similar to parallelo- gram FB, they are about the same diagonal [Prop. 6.26]. Let their (common) diagonal DB have been drawn, and let the (rest of the) figure have been described. Therefore, since (complement) CF is equal to (com- plement) FE [Prop. 1.43], and (parallelogram) FB is common, the whole (parallelogram) CH is thus equal to the whole (parallelogram) KE. But, (parallelogram) CH is equal to CG, since AC (is) also (equal) to CB [Prop. 6.1]. Thus, (parallelogram) GC is also equal to EK. Let (parallelogram) CF have been added to both. Thus, the whole (parallelogram) AF is equal to the gnomon LMN. Hence, parallelogram DB — that is to say, AD — is greater than parallelogram AF. Thus, for all parallelograms applied to the same straight-line, and falling short by a parallelogrammic figure similar, and similarly laid out, to the (parallelo- gram) described on half (the straight-line), the greatest is the [parallelogram] applied to half (the straight-line). (Which is) the very thing it was required to show. Proposition 28* To apply a parallelogram, equal to a given rectilin- ear figure, to a given straight-line, (the applied parallel- ogram) falling short by a parallelogrammic figure similar to a given (parallelogram) . It is necessary for the given rectilinear figure [to which it is required to apply an equal (parallelogram)] not to be greater than the (parallelo- gram) described on half (of the straight-line) and similar to the deficit. Let AB be the given straight-line, and C the given rectilinear figure to which the (parallelogram) applied to 185 ETOIXEIfiN 9'. ELEMENTS BOOK 6 Ttapd xf]v BoiDeToav eMelav xfjv AB xG SoiSevxi eO'duypajji^cp xG F laov TtapaXXr)X6ypajj.(jiov TtapapaXeTv cXXcTttov d'Ssi rca- paXX/]Xoypd|jijKp o^ioup ovxi xG A. 6 H O Z K Tex[if]OT}« f) AB 8[)(a xaxd xo E arjjieiov, xal dva- yeypacp'dco duo xfj? EB xG A o^toiov xal 6(ioiw<; xd^iervov xo EBZH, xal au^KETiXrjpwa'dw xo AH TtapaXXr)X6ypa^ov. EE \itv ouv iaov eaxl xo AH xG T, yeyovog av ei'r] xo era- xa)cdev 7tapapepXr]xai yap napa xf]v So-deTaav euiJeTav xf]v AB xG 8oi9evxi eui9uypd[i|jiw xG T laov TtapaXXrjXoypa^ov xo AH eXXeljtov eiBei 7tapaXX/]Xoypd|i^.tL> xG HB 6[io(cl> ovxi xG A. ei Se ou, [iziZov eaxto xo 0E xou T. 1'aov 8e xo 9E xG HB- [leiZov dpa xal xo HB xou T. G 8f] ^si£6v eaxi xo HB xou r, xauxr] xfj U7tepo)(fj 1'aov, xG 8e A ojioiov xal o^ioicx xei^ievov xo auxo auveaxdxo xo KAMN. dXXa xo A xG HB [eaxiv] o^toiov xal xo KM dpa xG HB eaxiv o^ioiov. eaxw ouv 6^i6Xoyo<; f] ^.ev KA xf) HE, f] 8e AM xfj HZ. xal snei laov eaxl xo HB xou; T, KM, ^.a^ov dpa eaxl xo HB xou KM- ^lei^wv dpa eaxl xal f) [lev HE xfj? KA, r] 8s HZ xfj? AM. xela-dw xrj ^tev KA Tar] f] HS, xfj 8e AM Xar] f] HO, xal aupi7i£7T;Xr]pGa , dw xo SHOn TtapaXXr)X6ypa^ov laov dpa xal o\±oiov eaxi [xo Hn] xG KM [dXXa xo KM xG HB 6\Loi6v eaxiv]. xal xo Hn dpa xG HB o^toiov eaxiv Ttepl xf]v auxrjv dpa Bid^texpov eaxi xo Hn xG HB. eax« auxGv Sid^texpoi; f] HnB, xal xaxayeypdcp^w xo a)(rjpc. 'EtieI ouv 'laov eaxl xo BH xoI<; T, KM, Gv xo Hn xG KM eaxiv iaov, Xoitcoc dpa 6 TX$ yvo^iwv XoitcG xG T laoc, eaxiv. xal CTtel 'laov eaxl xo OP xG SE, xoivov npoaxeia'do xo HB' 6Xov dpa xo OB oXw xG SB i'aov eaxiv. dXXa xo SB xG TE eaxiv laov, etcei xal rcXeupa f] AE TtXeupa xfj EB eaxiv \ar\- xal xo TE dpa xG OB eaxiv i'aov. xoivov Ttpoaxeia'dw xo SE - 6Xov dpa xo TE oXw xG $XT yvG^tovi saxiv laov. dXX' 6 <3>XT yvG^twv xG V zbziyfti} I'aoc xal xo TE dpa xG T eaxiv iaov. napa xf)v SoiJelaav dpa euiDeTav xf]v AB xG Soif)evxi eu-duypd^jjiw xG V i'aov TiapaXX^Xoypa^ov TiapapepXrjxai xo ET eXXeliiov eiSei 7iapaXXr]Xoypd^w xG nB 6[ioiw ovxi AB is required (to be) equal, [being] not greater than the (parallelogram) described on half of AB and similar to the deficit, and D the (parallelogram) to which the deficit is required (to be) similar. So it is required to apply a parallelogram, equal to the given rectilinear figure C, to the straight-line AB, falling short by a parallelogrammic figure which is similar to D. H G P F K N Let AB have been cut in half at point E [Prop. 1.10], and let (parallelogram) EBFG, (which is) similar, and similarly laid out, to (parallelogram) D, have been de- scribed on EB [Prop. 6.18]. And let parallelogram AG have been completed. Therefore, if AG is equal to C then the thing pre- scribed has happened. For a parallelogram AG, equal to the given rectilinear figure C, has been applied to the given straight-line AB, falling short by a parallelogram- mic figure GB which is similar to D. And if not, let HE be greater than C. And HE (is) equal to GB [Prop. 6.1]. Thus, GB (is) also greater than C. So, let (parallelo- gram) KLMN have been constructed (so as to be) both similar, and similarly laid out, to D, and equal to the ex- cess by which GB is greater than C [Prop. 6.25]. But, GB [is] similar to D. Thus, KM is also similar to GB [Prop. 6.21]. Therefore, let KL correspond to GE, and LM to GF. And since (parallelogram) GB is equal to (figure) C and (parallelogram) KM, GB is thus greater than KM. Thus, GE is also greater than KL, and GF than LM. Let GO be made equal to KL, and GP to LM [Prop. 1.3]. And let the parallelogram OGPQ have been completed. Thus, [GQ] is equal and similar to KM [but, KM is similar to GB] . Thus, GQ is also similar to GB [Prop. 6.21]. Thus, GQ and GB are about the same diag- onal [Prop. 6.26]. Let GQB be their (common) diagonal, and let the (remainder of the) figure have been described. Therefore, since BG is equal to C and KM, of which GQ is equal to KM, the remaining gnomon UWV is thus equal to the remainder C. And since (the complement) PR is equal to (the complement) OS [Prop. 1.43], let (parallelogram) QB have been added to both. Thus, the whole (parallelogram) PB is equal to the whole (par- 186 ETOIXEIfiN 9'. ELEMENTS BOOK 6 t<3 A [eTieiSyjTcep to IIB tw HII o^ioiov eaxiv]- oTtep eSei allelogram) OB. But, OB is equal to TT, since side noifjoai. AE is equal to side EB [Prop. 6.1]. Thus, TE is also equal to PB. Let (parallelogram) OS have been added to both. Thus, the whole (parallelogram) TS is equal to the gnomon VWU. But, gnomon VWU was shown (to be) equal to C. Therefore, (parallelogram) TS is also equal to (figure) C. Thus, the parallelogram ST, equal to the given rec- tilinear figure C, has been applied to the given straight- line AB, falling short by the parallelogrammic figure QB, which is similar to D [inasmuch as QB is similar to GQ [Prop. 6.24] ]. (Which is) the very thing it was required to do. t This proposition is a geometric solution of the quadratic equation x 2 — ax+/3 = 0. Here, x is the ratio of a side of the deficit to the corresponding side of figure D, a is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the deficit running along AB, and (3 is the ratio of the areas of figures C and D. The constraint corresponds to the condition < a 2 /A for the equation to have real roots. Only the smaller root of the equation is found. The larger root can be found by a similar method. ITocpa tt)V BcnMaav eru'deTocv tG BotJevti eufluYpd^cp 'laov TTapaXXrjXoypajujiov TtapapaXsIv OrapPaXXov ri'Bsi ua- pocXXr)Xoypd^uw 6\ioi(x) xw Bo'devTi. A M r \ / A E/ \ k / #Uo / / e N n e h "Eaxw f] y.ev Bo'rMaa su-vMa f] AB, to 8e 8oi9sv eu-OuYpa^ov, CS 5ei I'aov rcapa tt)v AB TtapapaXsIv, to T, S 8e Bel 6[ioiov UTteppdXXeiv, to A' 8eT 8f) rcapa tt]v AB sMeiav tw T eu^UYpd^iico laov 7i:apaXXr]X6Ypa|jijj.ov napa- PaXsiv UTieppdXXov ri'Bsi 7tapaXX/]XoYpd|ji^to ofioicp tc5 A. TeT[i.r)ai[)co f) AB B[)(a xaTa to E, xod dvayeYpd'&o duo t/)c EB to A ojioiov xal ojioimc; xri^iervov rcapaX- X/]XoYpa^tiov to BZ, xal auvapicpoTepoig ^tev toT<; BZ, T laov, to 8e A 6[ioiov xdi 6[ioia>c xsifisvov to ocuto au- vEOTaTW to H6. o^ioXoyoc; 8e eaTCO f) ^ev K6 Tfj ZA, f) 8e KH Tfj ZE. xdi ETte! ^£i£6v ecru to H0 toO ZB, ^d^wv dpa ecttI xdi f] ^lev K9 Tfj? ZA, f] 8s: KH Tfj ZE. expepXf]a , doaav di ZA, ZE, xdi Tfj [iev K0 for) soto f) ZAM, Tfj 8e KH for) f] ZEN, xdi au^7i£TtXr)pcfo , do to MN- to MN dpa iw H9 laov ts eaTi xal ojioiov. dXXa to H0 t£5 EA sgtiv o^oiov Proposition 29 f To apply a parallelogram, equal to a given rectilin- ear figure, to a given straight-line, (the applied parallelo- gram) overshooting by a parallelogrammic figure similar to a given (parallelogram) . L M H \ c A El x\fr, L / / IN N Q O G Let AB be the given straight-line, and C the given rectilinear figure to which the (parallelogram) applied to AB is required (to be) equal, and D the (parallelogram) to which the excess is required (to be) similar. So it is required to apply a parallelogram, equal to the given rec- tilinear figure C, to the given straight-line AB, overshoot- ing by a parallelogrammic figure similar to D. Let AB have been cut in half at (point) E [Prop. 1.10], and let the parallelogram BF, (which is) similar, and similarly laid out, to D, have been described on EB [Prop. 6.18]. And let (parallelogram) GH have been con- structed (so as to be) both similar, and similarly laid out, to D, and equal to the sum of BF and C [Prop. 6.25]. And let KH correspond to FL, and KG to FE. And since (parallelogram) GH is greater than (parallelogram) FB, 187 ETOIXEIfiN 9'. ELEMENTS BOOK 6 xal to MN dpa tG EA 6[ioi6v egtiv Ttepl ttjv auxrjv dpa 8id|jieTp6v eoxi to EA to MN. f^x'dw auTGv Bid^iETpog #) ZS, xai xaTayeYP&P'fl" TO a X^ a - 'EtteI Xaov sotI to H6 toTc; EA, T, dXXd to H6 tG MN iaov eotiv, xal to MN dpa toTc; EA, T Xaov eotlv. xoivov dcprjprp'dw to EA- Xoittoc; dpa 6 ^/X<i> Y^Gpiwv tG T sotiv I'oog. xal etieI iar) egtIv f) AE xfj EB, iaov eoti xal to AN tG NB, touteotl tG AO. xoivov TtpoaxEiadw to EE 1 oXov dpa to AS iaov eot ! tG yvGjiovi. dXXd 6 f&X^ yvG^icov tG r "taoc, eaTiv xal to AE dpa tG T Xaov eotiv. Ilapa Trjv ScdeTaav dpa sui&ETav tt]v AB tG BotSevti eru'duypafji^cp tG T Xaov KapaXXrjXoYpotji^ov TtapapspXrjTai to AS ujieppdXXov eiSei TCapaXXrjXoYpdjijKp tG nO ojioicp ovti tG A, etisI xal tG EA sotiv ojioiov to On- orap e6el Tioifjaai. KH is thus also greater than FL, and KG than FE. Let FL and FE 1 have been produced, and let FLM be (made) equal to KH, and FEN to ifG [Prop. 1.3]. And let (parallelogram) MN have been completed. Thus, MN is equal and similar to GH. But, GH is similar to EL. Thus, MJV is also similar to EL [Prop. 6.21]. EL is thus about the same diagonal as MN [Prop. 6.26]. Let their (common) diagonal FO have been drawn, and let the (remainder of the) figure have been described. And since (parallelogram) GH is equal to (parallel- ogram) EL and (figure) C, but GH is equal to (paral- lelogram) MN, MN is thus also equal to EL and C. Let EL have been subtracted from both. Thus, the re- maining gnomon XWV is equal to (figure) C. And since AE is equal to EB, (parallelogram) AN is also equal to (parallelogram) NB [Prop. 6.1], that is to say, (parallel- ogram) LP [Prop. 1.43]. Let (parallelogram) EO have been added to both. Thus, the whole (parallelogram) AO is equal to the gnomon VWX. But, the gnomon VWX is equal to (figure) C. Thus, (parallelogram) AO is also equal to (figure) C. Thus, the parallelogram AO, equal to the given rec- tilinear figure C, has been applied to the given straight- line AB, overshooting by the parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [Prop. 6.24] . (Which is) the very thing it was required to do. t This proposition is a geometric solution of the quadratic equation x 2 +a x — (3 = 0. Here, x is the ratio of a side of the excess to the corresponding side of figure D, a is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the excess running along AB, and (3 is the ratio of the areas of figures C and D. Only the positive root of the equation is found. X' Proposition 30+ Trjv SoiMoav sudslav TteTCpaa|jievr]v dxpov xal (isaov To cut a given finite straight-line in extreme and mean Xoyov te^eTv. ratio. r z o H A J B J B 'A TEotw f) So-dsTaa EU-dsTa K£7i£paa[i£vr) f] AB- 5eT 8f] ttjv AB EtrdsTav dxpov xal ^saov Xoyov te^eiv. D Let AB be the given finite straight-line. So it is re- quired to cut the straight-line AB in extreme and mean 188 ETOIXEIfiN 9'. ELEMENTS BOOK 6 AvayeypdcpiDw duo xfj<; AB xexpdyiovov to Br, xal rca- paPepXr]o , dw napd xr)v Ar xco Br iaov 7iapaXXr]X6ypa[j.(jiov xo TA (meppdXXov el'8ei xw AA o^xoicp iw Br. Texpdycovov 8e eaxi xo Br- xexpdywvov dpa eaxi xdi xo AA. xdi £7td laov eaxi xo Br xw TA, xoivov dcp/]pr)aiL>M xo TE- Xomov dpa xo BZ XoiraS xcp AA eaxiv '(aov. eaxi 8s auxw xdi laoywviov xfiv BZ, AA dpa avxiTiCKovdaaiv ai TiXeupal ai nepl xd<; iaa<; ytoviac eaxiv dpa ci><; f) ZE -rcpoc; xr]v EA, ouxwc; f) AE npo? xrjv EB. Tar) Se f] ^iev ZE xrj AB, f] 8e EA xfj AE. eaxiv dpa &>c, rj BA rcpoc; x/]v AE, ouxgx f) AE Ttpoc; xfjv EB. ^iei£«v 5e f] AB xrjc; AE- ^ei^tov dpa xal r) AE xfjc; EB. H dpa AB euiDeTa dxpov xal ^.eaov Xoyov xex^irjxai xaxd xo E, xal xo ^.elCov auxrj<; x[ifj^.d eaxi xo AE- oitep e8ei Ttoirjaai. ratio. Let the square BC have been described on AB [Prop. 1.46], and let the parallelogram CD, equal to BC, have been applied to AC, overshooting by the figure AD (which is) similar to BC [Prop. 6.29]. And BC is a square. Thus, AD is also a square. And since BC is equal to CD, let (rectangle) CE have been subtracted from both. Thus, the remaining (rect- angle) BF is equal to the remaining (square) AD. And it is also equiangular to it. Thus, the sides of BF and AD about the equal angles are reciprocally proportional [Prop. 6.14]. Thus, as FE is to ED, so AE (is) to EB. And FE (is) equal to AB, and ED to AE. Thus, as BA is to AE, so AE (is) to EB. And AB (is) greater than AE. Thus, AE (is) also greater than EB [Prop. 5.14]. Thus, the straight-line AB has been cut in extreme and mean ratio at E, and AE is its greater piece. (Which is) the very thing it was required to do. t This method of cutting a straight-line is sometimes called the "Golden Section" — see Prop. 2.11. Xa'. 'Ev xou; opiJoycovioic; xpiyiovoic; xo duo xfjc; xr)v op'drjv ywviav U7ioxeivouar]<; TtXeupac; eI8o<; I'aov eaxi xou; duo xwv xr]v 6pi}r]v ywviav Ttepiexouawv TtXeupov ei5eai xolg ojioioic; xe xal o^oicoc; dvaypacpouevoic;. 'Eaxw xpiywvov op-doywviov xo ABr op'drjv e)(ov xrjv Otio BAT yoviav Xeyw, oxi xo anb xfjc BT eTSoc 'laov eaxi xolc; duo xfiv BA, Ar eToeai xoI<; ouoiok; xe xal 6uo(w<; dvaypacpo^ievoic;. "H)(Tf)w xd'dexoc; f) AA. 'Etcei oov ev opiDoywviw xpiywvw ifi ABr anb xrjc npbc, xcp A opi^c; yoviac; era xrjv Br pdaiv xd-fJexoc; rjxxai f\ AA, xd ABA, AAr Ttpo? xfj xaiSexcp xpiywva opioid eaxi x£> xe 0X0 xw ABr xal dXXr)Xou;. xal enel o^ioiov eaxi xo ABr xS ABA, eaxiv dpa cb<; f) TB 7tp6<; xfjv BA, ouxwc; f) AB npo? xr]v BA. xal CTiel xpeTc eui9eTai dvdXoyov eiaiv, eaxiv cbc; f) npd>xr) 7ip6<; xrjv xpixrjv, ouxw<; xo duo xrjc; 7tpcoxr]C eTBoc Tipoc Proposition 31 In right-angled triangles, the figure (drawn) on the side subtending the right-angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right-angle. Let ABC be a right-angled triangle having the angle BAC a right-angle. I say that the figure (drawn) on BC is equal to the (sum of the) similar, and similarly described, figures on BA and AC. Let the perpendicular AD have been drawn [Prop. 1.12]. Therefore, since, in the right-angled triangle ABC, the (straight-line) AD has been drawn from the right- angle at A perpendicular to the base BC, the trian- gles ABD and ADC about the perpendicular are sim- ilar to the whole (triangle) ABC, and to one another [Prop. 6.8]. And since ABC is similar to ABD, thus 189 ETOIXEIfiN 9'. ELEMENTS BOOK 6 to duo xfjg Bsuxspag to ojioiov xal o^ioicoc; dvaypacpojisvov. (be; dpa f) TB rcpoc; xrjv BA, ouxwc; xo duo xfjc; TB eT8oc; Tipog xo duo xfjc BA xo ojioiov xdi ojioiwc; dvaypacpo^iEvov. 8ia xd auxd 6f) xdi (be; f) Br Tipoc; xrjv TA, ouxcog xo drco xrj<; Br sTBoc; Ttpoc; xo duo xfjc TA. <baxe xdi (be; f) Br upog xdg BA, Ar, oux(oc; xo arco xrjg Br eTSoc; upoc; xd duo x£5v BA, Ar xd ojjioia xdi ojioitog dvaypacpojieva. Tar] Bs rj Br xdu; BA, Ar- Taov dpa xdi xo duo xfjc Br eiSog xou; duo xwv BA, Ar elbeai xou; ojioioic; xs xdi o^iolioc; dvaypacpo^ievoic;. 'Ev dpa xolg op'doywvioic; xpiywvou; xo a.nb xfjc xrjv opiSrjv ywviav unoxeivouarjc; TtXcupdc; eT6oc; Taov eaxl xou; dTio xwv xrjv opi^rjv yioviav TiepiexouaSSv nXeupfiv ri'Bsai xoTg o^oiou; xs xal o^oiwg dvaypacpojis vou;- onep eBei 8eic;ai. X(3'. 'Edv 80o xpiyova auvxrdfj xaxd (iiav ywviav xdg 860 TtXeupdc; xdu; 8ual TtXeupau; dvdXoyov £)(ovxa (baxs xdg opioXoyouc; otuxwv TtXsupdc xal 7iapaXXr]Xou<; elvai, ai Xomal xwv xpiywvwv TiXeupal z% sundae; eaovxai. A Br e 'Eaxaj 860 xpiywva xd ABr, ArE xd? 860 rcXeupdc xdc BA, Ar xdu; 8ual rcXeupau; xdu; Ar, AE dvdXoyov s/ovxa, (be ^iev xrjv AB Tip6<; xrjv Ar, oux«<; xrjv Ar Ttpoc; xrjv AE, TiapdXXrjXov 8s xrjv (iev AB xrj Ar, xrjv 8e Ar xfj AE- Xsyco, oxi £7t' sO'dslac saxlv f] Br xfj TE. 'Etce! yap napdXXrjXoc; eaxiv fj AB xrj Ar, xal tic, auxd? ejUTtSTtxwxev eu'deTa fj AT, ai evaXXd<; ywviai ai bnb BAT, ATA i'aai dXXrjXaic; riaiv. 81a xd auxd 8fj xal fj \mb TAE xfj O716 ABA iarj saxiv. (benx xal fj 6716 BAr xrj utco TAE iaxiv larj. xal ETiel 860 xpiywvd eaxi xd ABr, ArE puav ywviav xfjv Tipoc; xw A [iia ywvia xfj rcpoc xo A larjv s^ovxa, Tiepl as CB is to BA, so ,45 (is) to BD [Def. 6.1]. And since three straight-lines are proportional, as the first is to the third, so the figure (drawn) on the first is to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.]. Thus, as CB (is) to BD, so the fig- ure (drawn) on CB (is) to the similar, and similarly de- scribed, (figure) on BA. And so, for the same (reasons), as BC (is) to CD, so the figure (drawn) on BC (is) to the (figure) on CA. Hence, also, as BC (is) to BD and DC, so the figure (drawn) on BC (is) to the (sum of the) similar, and similarly described, (figures) on BA and AC [Prop. 5.24]. And BC is equal to BD and DC. Thus, the figure (drawn) on BC (is) also equal to the (sum of the) similar, and similarly described, figures on BA and AC [Prop. 5.9]. Thus, in right-angled triangles, the figure (drawn) on the side subtending the right-angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right-angle. (Which is) the very thing it was required to show. Proposition 32 If two triangles, having two sides proportional to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straight-on (with respect to one another). D B C E Let ABC and DCE be two triangles having the two sides BA and AC proportional to the two sides DC and DE—so that as AB (is) to AC, so DC (is) to DE— and (having side) AB parallel to DC, and AC to DE. I say that (side) BC is straight-on to CE. For since AB is parallel to DC, and the straight-line AC has fallen across them, the alternate angles BAC and ACD are equal to one another [Prop. 1.29]. So, for the same (reasons), CDE is also equal to ACD. And, hence, BAC is equal to CDE. And since ABC and DCE are two triangles having the one angle at A equal to the one 190 ETOIXEIfiN 9'. ELEMENTS BOOK 6 8s tolc, XaoLc, ycoviac; Tag TtXeupac; dvdXoyov, toe; xr)V BA Ttpoc; ttjv Ar, outoc; ttjv TA Ttpoc; xr)v AE, iaoycjviov apa eoxl to ABr Tpiycovov tw ArE Tpiywvw- for) apa f] utio ABr ywvia Tfj utio ArE. e8eix'dr) 8s xai f] utio ArA Tfj utco BAr for)- oXr] apa f] utio ArE 8uai TaT? utio ABr, BAr for) ecrciv. xoivf] Tipoaxeicydto f] utio ATB- ai apa utio ArE, ArB toic utio BAr, ArB, TBA foai slaw. dXX" ai utio BAr, ABr, ArB Suoiv op'ddic; foai eioiv xal ai utio ArE, ArB apa 8uaiv op'ddic; foai siaiv. Tipoc; 8f] tivi cu-deia Tfj Ar xai iw Tcp6<; auTfj ar^eiw iu T 8uo eu-delai ai Br, TE [irj etc! t& auTa \iepr) xei^tevai iac, ecpec;rj<; ™c, utco ArE, ArB 8uaiv op'ddic; foac; TioioOaiv etc' eu'deiac; apa ecrciv f] Br Tfj rE. °Edv apa 8uo Tpiycova auvTcdfj xaTa ^.iav ywviav Tag 8uo TiXeupdc Talc; Suai TtXeupau; dvdXoyov exovTa wots tolc, o^ioXoyouc; auTWv TtXeupac; xai TiapaXXr)Xou<; elvai, ai XoiTial twv Tpiy«v«v TtXeupai etc' eu'deiac; eaovTai- oTtep eBei BeT^ai. Xy'. 'Ev toT<; Took; xuxXoic; ai ywviai tov aikov e)(0UGi Xoyov tocT<; Ttepicpepeiaic;, ecp'' Sv peprjxaaiv, edv tc Tipoc; tou; xevTpoic; edv tc Ttpoc; tolc Ttepicpepeiaic; Sai pepr)xuiai. 'Ecrccoaav fooi xuxXoi oi ABr, AEZ, xal Ttpoc; ^ev to!<; xevTpoic auTGv toTc; H, ywviai eoiuoav ai utco BBT, E6Z, Ttpoc 5e Talc; Tiepicpepeiau; ai utio BAr, EAZ- Xeyco, oti ecrciv cbc; f) BV Ttepicpepeia Ttpoc; t/]v EZ Tiepicpepeiav, outox fj tc utio BHT yovia Ttpoc; t/)v utio E6Z xai f) utco BAr Ttpoc; tt)v utio EAZ. Keicydtoaav yap Tfj [ie\i BT Ttepicpepeia foai xaTa to ec;rjc; 6aai8r)TtoTouv ai TK, KA, Tfj 8e EZ Tiepicpepeia foai oaai- 8r)TtoTouv ai ZM, MN, xai CTteCeux'dwaav ai HK, HA, 0M, 6N. 'Etcei ouv i'aai eiaiv ai Br, TK, KA Tiepicpepeiai dXXf|Xaic;, foai eiai xai ai utio BBT, THK, KHA ywviai dXXrjXaic oaaTtXaaiov apa ecrciv f] BA Ttepicpepeia xfj? Br, ToaauTa- TtXaaiwv ecrci xai f] utio BHA yuvia Tfj? utio BHr. 8id Ta angle at D, and the sides about the equal angles pro- portional, (so that) as BA (is) to AC, so CD (is) to DE, triangle ABC is thus equiangular to triangle DCE [Prop. 6.6]. Thus, angle ABC is equal to DCE. And (an- gle) ACD was also shown (to be) equal to BAC. Thus, the whole (angle) ACE is equal to the two (angles) ABC and BAC. Let ACB have been added to both. Thus, ACE and ACB are equal to BAC, ACB, and CBA. But, BAC, ABC, and ACB are equal to two right-angles [Prop. 1.32]. Thus, ACE and ACB are also equal to two right-angles. Thus, the two straight-lines BC and CE, not lying on the same side, make adjacent angles ACE and ACB (whose sum is) equal to two right-angles with some straight-line AC, at the point C on it. Thus, BC is straight-on to CE [Prop. 1.14]. Thus, if two triangles, having two sides proportional to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straight-on (with respect to one another). (Which is) the very thing it was required to show Proposition 33 In equal circles, angles have the same ratio as the (ra- tio of the) circumferences on which they stand, whether they are standing at the centers (of the circles) or at the circumferences. Let ABC and DEF be equal circles, and let BGC and EHF be angles at their centers, G and H (respectively), and BAC and EDF (angles) at their circumferences. I say that as circumference BC is to circumference EF, so angle BGC (is) to EHF, and (angle) BAC to EDF. For let any number whatsoever of consecutive (cir- cumferences), CK and KL, be made equal to circumfer- ence BC, and any number whatsoever, FM and MN, to circumference EF. And let GK, GL, HM, and HN have been joined. Therefore, since circumferences BC, CK, and KL are equal to one another, angles BGC, CGK, and KGL are also equal to one another [Prop. 3.27]. Thus, as many times as circumference BL is (divisible) by BC, so many 191 ETOIXEIfiN 9'. ELEMENTS BOOK 6 auxd 8f] xal oaaTcXaatav eaxlv rj NE Tcepicpepeia xfjc EZ, xo- aauxaTcXaalov eaxl xal f] utco N0E ytovia xfj? utco E0Z. si apa for) eaxlv f] BA Tcepicpepeia xfj EN Tcepicpepeia, !ar] eaxl xal ycovia f\ utco BHA xfj utio EON, xal el [iei^tov eaxlv f] BA Tiepicpepeia xfj? EN Tcepicpepeia?, jiei^Mv eaxl xal f) utio BHA ycovla xfj? utio EON, xal el eXdaawv, eXdaawv. xeaadpwv 8f| ovxov jieye-dfiiv, Suo (lev itepicpepeiwv xfiv Br, EZ, 660 8e ywvifiv x£Sv utio BBT, E6Z, eTXr)Tixai xfj? ^iev Br Tiepi- cpepeia? xal xfjt; utco BHr ycovia? ladxi? TcoXXaTcXaalwv fj xe BA Tcepicpepeia xal r\ utco BHA y«via, xfj? 8e EZ Tcepicpepeia? xal xfj? utco E0Z ywvia? fj xe EN Tcepicpepia xal f\ utco E9N ycovia. xal 8e8eixxai, 6x1 el UTcepe^ei f\ BA Tcepicpepeia xfj? EN Tcepicpepeia?, UTcepexei xal rj utco BHA ytovla xfj? utco EON ywvla?, xal el Tar), Tar), xal el eXdaawv, eXdaawv. eaxiv apa, cb? f] Br Tcepicpepeia Tcpo? xf]v EZ, ouxco? f] utco BHr ywvia Tcpo? xf]V utco EOZ. dXX' <i>? f] utco BHr ywvia Tcpo? xfjv utco EOZ, ouxo? f\ utco BAr Tcpo? xf)V utco EAZ. 8iTcXaa(a yap exaxepa exaxepa?. xal 6? apa f] Br Tcepicpepeia Tcpo? xf]V EZ Tcepicpepeiav, ouxw? fj xe utco BHr yovia Tcpo? xf]v utco E0Z xal f] utco BAr Tcpo? xf)v utco EAZ. 'Ev apa xol? Taoi? xuxXoi? ai ywviai xov auxov exouai Xoyov xal? Tcepicpepelai?, ecp' Sv pepfjxaaiv, edv xe Tcpo? xoT? xevxpoi? edv xe Tcpo? xal? Tcepicpepelai? c5ai peprjxuTai - oTcep e8ei 8eT<;ai. t This is a straight-forward generalization of Prop. 3.27 times is angle BGL also (divisible) by BGC. And so, for the same (reasons), as many times as circumference NE is (divisible) by EF, so many times is angle NHE also (divisible) by EHF. Thus, if circumference BL is equal to circumference EN then angle BGL is also equal to EHN [Prop. 3.27], and if circumference BL is greater than circumference EN then angle BGL is also greater than EHN J and if (BL is) less (than EN then BGL is also) less (than EHN). So there are four magnitudes, two circumferences BC and EF, and two angles BGC and EHF. And equal multiples have been taken of cir- cumference BC and angle BGC, (namely) circumference BL and angle BGL, and of circumference EF and an- gle EHF, (namely) circumference EN and angle EHN. And it has been shown that if circumference BL exceeds circumference EN then angle BGL also exceeds angle EHN, and if (BL is) equal (to EN then BGL is also) equal (to EHN), and if (BL is) less (than EN then BGL is also) less (than EHN). Thus, as circumference BC (is) to EF, so angle BGG (is) to EHF [Dei. 5.5]. But as angle BGC (is) to EHF, so (angle) BAG (is) to EDF [Prop. 5.15]. For the former (are) double the latter (re- spectively) [Prop. 3.20]. Thus, also, as circumference BC (is) to circumference EF, so angle BGC (is) to EHF, and BAG to EDF. Thus, in equal circles, angles have the same ratio as the (ratio of the) circumferences on which they stand, whether they are standing at the centers (of the circles) or at the circumferences. (Which is) the very thing it was required to show. 192 ELEMENTS BOOK 7 Elementary Number Theory^ tThe propositions contained in Books 7-9 are generally attributed to the school of Pythagoras. 193 ETOIXEIfiN C- ELEMENTS BOOK 7 "Opoi. a'. Movdc; saxiv, xai9' fjv sxaaxov x65v ovxcov §v XiyeTw.. P'. Api/d^ioc; 8s to ex jiovd8«v auyxei|jievov TiXfj'dot;. y'. Mepog eaxlv dpi/d^ioc; dpi/d^ioO 6 eXdaawv xou (lei^ovot;, oxav xaxajiexpfj xov [isic^ova. 8'. Mspr) 5s, oxav y.r\ xaxajisxp/j. z. noXXaTiXdaiog Ss 6 [isi^cov xou sXdaaovoc;, oxav xa- xajjiexpfjxai Otto xou eXdaaovoc;. 'Apxioc; dpi%6c; soxlv 6 Si/a Siaipoujievog. C. Ilepiaaoc; 8e 6 fjirj Biaipou^evot; Si^a f\ [6] jiovdBi Siacpspwv dpxiou dpidjjioO. /)'. Apxidxic; apxioc; dpi%6c; eaxiv 6 utto dpxbu dpidjiou jiexpoujievo^ xaxd dpxiov dpidjiov. 'Apxidxic; 8e Ttepiaaoc; eaxiv 6 uuo dpxiou dpid[iou jjiexpoujievog xaxd itspiaaov dpiiSjiov. i'. Ilspiaadxic; 8s Ttepiaaoc; dpi%6c; eaxiv 6 Otco Ttepiaaou dpiiSpiou ^expou^ievoc; xaxd nepiaaov dpid^iov. ia'. IlpGxoc; dpid^ioc; eaxiv 6 [iovd8i \i6vx] ^.expou^tevoc;. ip'. ilpwxoi icpoc; dXXf]Xou<; dpid^oi eiaiv oi ^tovdSi [i6vr] ^lexpou^tevoi xoivG ^texpw. iy'. Suvdexoc; dpi'd^oc; eaxiv 6 apidjiw xivi fjiexpou^ievoc;. 18'. £uv$exoi 8e upoc; dXXf]Xou<; dpi-dnoi eiaiv oi dpi%« xivi ^lexpou^ievoi xoivfi ^.expw. ie'. Aprdjioc; dprd^iov TcoXXanXaaid^eiv XcYexai, oxav, oaai elolv ev auxw piovd8ec;, xoaauxdxic; auvxei}/] 6 ttoX- XajcXaaia^o^ievoc;, xal xe\ir\Ta.i xu;. if'. "Oxav 8e Suo dpi'&[Jiol TtoXXajcXaaidaavxec; dXXr|Xouc; jcoiwai xiva, 6 yevojjlevo^ eraTceBoc; xaXeTxai, nXeupal Be: auxou oi TcoXXaiiXaaidaavxec; dXXr|Xouc; dpid^ioi. iC- "Oxav 8e xpeTc; dpid^iol TcoXXaiiXaaidaavxec; dXXr|Xouc; Ttoifiai xiva, 6 Yevo^evog axepeoc; eaxiv, nXeupal Se auxou oi TroXXaTtXaaidaavxec; dXXr|Xouc; dpidjioL irj'. TexpdY«vog dpii9[i6c eaxiv 6 iadxic; i'aoc; f\ [6] uno Suo Taiov dpidjifiv Ttepie/oijievoc;. it}'. Kupog 8s 6 iadxic; laog iadxic; fj [6] utto xpifiv lacov dpii9|jL«v Tiepie)(6^evoc;. x'. Apiduoi dvdXoYov eiaiv, oxav 6 Ttpwxoc; xou 8euxepou xal 6 xpixoc; xou xexdpxou iadxic; rj TtoXXaTiXdaioc; fj xo auxo (icpoc; rj xa auxd fiepr) waiv. xa'. "O[ioioi eixLTxeBoL xai axepeoi dpid^ioi eiaiv oi avdXoYov £)(ovx£<; xa<; TiXeupdt;. xp'. TeXsioc dpi-djiot; eaxiv 6 xolz eauxou (iepeaiv laoc, Definitions 1. A unit is (that) according to which each existing (thing) is said (to be) one. 2. And a number (is) a multitude composed of units. t 3. A number is part of a(nother) number, the lesser of the greater, when it measures the greater. * 4. But (the lesser is) parts (of the greater) when it does not measure it. § 5. And the greater (number is) a multiple of the lesser when it is measured by the lesser. 6. An even number is one (which can be) divided in half. 7. And an odd number is one (which can) not (be) divided in half, or which differs from an even number by a unit. 8. An even-times-even number is one (which is) mea- sured by an even number according to an even number. ^ 9. And an even-times-odd number is one (which is) measured by an even number according to an odd number.* 10. And an odd-times-odd number is one (which is) measured by an odd number according to an odd number. 8 11. A primed number is one (which is) measured by a unit alone. 12. Numbers prime to one another are those (which are) measured by a unit alone as a common measure. 13. A composite number is one (which is) measured by some number. 14. And numbers composite to one another are those (which are) measured by some number as a common measure. 15. A number is said to multiply a(nother) number when the (number being) multiplied is added (to itself) as many times as there are units in the former (number), and (thereby) some (other number) is produced. 16. And when two numbers multiplying one another make some (other number) then the (number so) cre- ated is called plane, and its sides (are) the numbers which multiply one another. 17. And when three numbers multiplying one another make some (other number) then the (number so) created is (called) solid, and its sides (are) the numbers which multiply one another. 18. A square number is an equal times an equal, or (a plane number) contained by two equal numbers. 19. And a cube (number) is an equal times an equal times an equal, or (a solid number) contained by three equal numbers. 194 ETOIXEIfiN C- ELEMENTS BOOK 7 20. Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third (is) of the fourth. 21. Similar plane and solid numbers are those having proportional sides. 22. A perfect number is that which is equal to its own parts tt t In other words, a "number" is a positive integer greater than unity. t In other words, a number a is part of another number b if there exists some number n such that na = b. § In other words, a number a is parts of another number b (where a < b) if there exist distinct numbers, m and n, such that na = mb. ' In other words, an even-times-even number is the product of two even numbers. * In other words, an even-times-odd number is the product of an even and an odd number. $ In other words, an odd-times-odd number is the product of two odd numbers. II Literally, "first". tt In other words, a perfect number is equal to the sum of its own factors. Auo dpid^cov dviatov exxeijievwv, dvducpaipoujisvou 8s dsl xou sXdaaovoc; oltzo xou [idCovoc;, sav 6 Xsmo^isvoc; ^r)Ss7toxs xaxajisxpfj xov npo sauxou, scoc oo Aeicp-drj jiovdc;, ol eZ, dp/rjc; dpi%ol TtpGxoi Ttpoc; dXXrjXouc; saovxai. At Proposition 1 Two unequal numbers (being) laid down, and the lesser being continually subtracted, in turn, from the greater, if the remainder never measures the (number) preceding it, until a unit remains, then the original num- bers will be prime to one another. B 1 r H F - H tC G Auo yap [dv(o«v] dpidjjicov xfiv AB, TA dvducpai- poujisvou del xou eXdaaovoc; dno xou jisi^ovoc; 6 Xsittojjievoc; (jirjBsTtoxs xaxa^sxpsixc.) xov npo sauxou, sex; ou Xsicp-dfj (iovdc Xeyw, oxi ol AB, TA TtpGxoi Ttpoc; dXXf]Xou<; slaiv, xouxsaxiv oxi xouc; AB, TA \xovclq \io\>r\ (jisxpsl. El ydp [ir\ siaiv ol AB, TA npfixoi npoc; dXXrjXouc;, ^sxpiqasi tic; auxouc; dpi'd^oc;. [isxpeixw, xdi eaxw 6 E- xal 6 [isv TA xov BZ ^isxpwv Xeittetio sauTou sXdaaova xov ZA, 6 8s AZ xov AH ^.STpwv Xsitcstco sauTou eXdaaova xov HT, 6 8s HT xov Z9 ^texpwv Xsitistw ^tovdSa xf]v 9A. Tksl ouv 6 E xov TA jiexpel, 6 8s TA xov BZ uexpel, xdi 6 E dpa xov BZ ^.expel' jiexpel 8s xdi oXov xov BA' xal Xoitcov dpa xov AZ [leiprpei.. 6 8s AZ xov AH ^isxpsT' xal 6 E dpa xov AH jisxpsT' [isxpsT 8s xal oXov xov AT- xal Xoittov dpa xov TH ^sxprjasi. 6 8s TH xov Z6 jisxpsl- B 1 1 D For two [unequal] numbers, AB and CD, the lesser being continually subtracted, in turn, from the greater, let the remainder never measure the (number) preceding it, until a unit remains. I say that AB and CD are prime to one another — that is to say, that a unit alone measures (both) AB and CD. For if AB and CD are not prime to one another then some number will measure them. Let (some number) measure them, and let it be E. And let CD measuring BF leave FA less than itself, and let AF measuring DC leave GC less than itself, and let GC measuring FH leave a unit, HA. In fact, since E measures CD, and CD measures BF, E thus also measures BF.^ And (E) also measures the whole of BA. Thus, (E) will also measure the remainder 195 ETOIXEIfiN C- ELEMENTS BOOK 7 xal 6 E dpa xov Z6 jiexpeT- (jiexpel Be xal oXov xov ZA- ^li* 1 .* And AF measures DG. Thus, _E also measures DG. xal Xoikt]v apa xrjv A0 piovdBa ^exprjaei dpi%6<; wv oTtep And (i?) also measures the whole of DC. Thus, (E) will eaxlv dSuvaxov. oux dpa xou<; AB, TA dpi%ou<; {LSTpfpei also measure the remainder CG. And CG measures FH. xic dpi%6c oi AB, TA dpa npoxoi Kpoc dXXr]Xou<; eiaiv Thus, _E also measures FiJ. And also measures the oiiep eSei SeT^ai. whole of FA. Thus, (i?) will also measure the remaining unit AH, (despite) being a number. The very thing is impossible. Thus, some number does not measure (both) the numbers AB and CD. Thus, AB and CD are prime to one another. (Which is) the very thing it was required to show. t Here, use is made of the unstated common notion that if a measures b, and b measures c, then a also measures c, where all symbols denote numbers. t Here, use is made of the unstated common notion that if a measures b, and a measures part of b, then a also measures the remainder of b, where all symbols denote numbers. P'- Auo dpnJ^icov Bo'devxov \xr\ upcoxcov upoc; dXXr)Xouc; xo tieyiaxov auxov xoivov jiexpov eupelv. A E T B 1 T r z H 1 A 1 "Eaxwaav oi Scedevxe^ Suo dprd^oi ^ir) npcoxoi Ttpoc dXXiqXouc; oi AB, TA. 8s! Br) x«v AB, TA xo ^eyiaxov xoivov (jiexpov eupe1v. Ei y.sv ouv 6 TA xov AB ^expet, ^expel Be xal eauxov, 6 TA dpa xwv TA, AB xoivov ^Jtexpov eaxiv. xal cpavepov, oxi xai ^iiyiaxov ouBelc; yap ^eiCcov xou TA xov TA \LSTpr\oe\.. Ei Be ou [icxpel 6 TA xov AB, xwv AB, TA dvducpai- poujievou del xou eXdaaovo<; olko xou ^.eiCovoc; Xeicp , dr]aexai uc, dpid^toc;, bz \xexpr\oei xov upo eauxou. ^.ovac fiev yap ou Xeicp-driaexar ei Be (jltq, eaovxai oi AB, TA Kpwxoi 7ip6<; dXXrjXouc ouep oux urcoxeixai. XeicpiSriaexai tic; dpa dpid^icx;, be, ^exprjoei xov Tipo eauxou. xai 6 \iev EA xov BE jiexpwv Xeiitexw eauxou eXdaaova xov EA, 6 Be EA xov AZ ^.expwv Xentexco eauxou eXdaaova xov ZT, 6 Be TZ xov AE [lexpeixw. euel ouv 6 TZ xov AE ^icxpel, 6 Be AE xov AZ ^lexpeT, xal 6 TZ apa xov AZ ^lexprjaei. [icxpel Be xal eauxov xal oXov dpa xov TA (jiexprpei. 6 Be TA xov BE ^exper xal 6 TZ dpa xov BE ^texpeT- [icxpel Be xal xov EA- xal oXov apa xov BA [iexpr|aer ^texpeT Be xal xov TA- 6 TZ apa xou<; AB, EA ^texpeT. 6 TZ dpa xwv AB, TA xoivov Proposition 2 To find the greatest common measure of two given numbers (which are) not prime to one another. A E - tC F Let AB and CD be the two given numbers (which are) not prime to one another. So it is required to find the greatest common measure of AB and CD. In fact, if CD measures AB, CD is thus a common measure oi CD and AB, (since CD) also measures itself. And (it is) manifest that (it is) also the greatest (com- mon measure). For nothing greater than CD can mea- sure CD. But if CD does not measure AB then some number will remain from AB and CD, the lesser being contin- ually subtracted, in turn, from the greater, which will measure the (number) preceding it. For a unit will not be left. But if not, AB and CD will be prime to one another [Prop. 7.1]. The very opposite thing was assumed. Thus, some number will remain which will measure the (num- ber) preceding it. And let CD measuring BE leave EA less than itself, and let EA measuring DF leave FC less than itself, and let CF measure AE. Therefore, since CF measures AE, and AE measures DF, CF will thus also measure DF. And it also measures itself. Thus, it will 196 ETOIXEIfiN C- ELEMENTS BOOK 7 [iexpov eaxiv. Xeyw 8r|, oxi xal [ieyiaxov. si yap [jlttj eaxiv 6 rZ xfiv AB, TA [ieyiaxov xoivov [iexpov, [iexprpei tic; xouc; AB, TA dpii9^ouc; dpid[i6c; [iei^iov «v xou TZ. [iexpeixw, xal cgxgj 6 H. xal eitel 6 H xov TA [iexpeT, 6 6e TA xov BE [iexpel, xal 6 H dpa xov BE [iexpeT- [iexpeT 8e xal oXov xov BA- xal Xoinov apa xov AE [iexprjaei. 6 Se AE xov AZ [iexpeT- xal 6 H apa xov AZ [iexprpei- [iexpeT 8s xal oXov xov AE xal Xoittov apa xov TZ [iexpiqaei 6 [ieii^cov xov eXdaaova- 6%ep eaxlv dSuvaxov oux dpa xouc; AB, TA dpn5[iouc; dpidjioc; xu; [iexprpei [iei^wv wv xou TZ- 6 TZ apa xwv AB, TA [ieyiaxov £° Tt xoivov (iexpov [onep eSel Belial]. also measure the whole of CD. And CD measures BE. Thus, CF also measures BE. And it also measures EA. Thus, it will also measure the whole of BA. And it also measures CD. Thus, CF measures (both) AB and CD. Thus, CF is a common measure of AB and CD. So I say that (it is) also the greatest (common measure). For if CF is not the greatest common measure of AB and CD then some number which is greater than CF will mea- sure the numbers AB and CD. Let it (so) measure (AB and CD), and let it be G. And since G measures CD, and CD measures BE, G thus also measures BE. And it also measures the whole of BA. Thus, it will also mea- sure the remainder AE. And AE measures DF. Thus, G will also measure DF. And it also measures the whole of DC. Thus, it will also measure the remainder CF, the greater (measuring) the lesser. The very thing is im- possible. Thus, some number which is greater than CF cannot measure the numbers AB and CD. Thus, CF is the greatest common measure of AB and CD. [(Which is) the very thing it was required to show] . I16pia[jia. 'Ex 8r) xouxou cpavepov, oxi edv dpid[i6c; 8uo dpid[iouc; [iexprj, xal xo [ieyiaxov auxSSv xoivov [iexpov [iexprpei- ojtep e8ei 8eT^ai. Corollary So it is manifest, from this, that if a number measures two numbers then it will also measure their greatest com- mon measure. (Which is) the very thing it was required to show. T • Tpiwv dpid[ic5v So'devxwv [irj Ttpwxcov izpoc, dXXr|Xouc; to [ieyiaxov auxov xoivov [iexpov eupeTv. Proposition 3 To find the greatest common measure of three given numbers (which are) not prime to one another. A B T A E Z TCaxoaav oi Bo-devxec; xpeT? dpn9[iol ]if\ Tipoxoi TCpoc dXXf|Xouc; oi A, B, E 5eT 8rj xov A, B, T xo [ieyiaxov xoivov [iexpov eupeTv. EiXr](p'd« yap Suo xwv A, B xo [ieyiaxov xoivov [iexpov 6 A- 6 8r) A xov T rjxoi [iexpeT rj ou [iexpeT. [iexpeixw itpoxepov [iexpeT Se xal xouc; A, B- 6 A apa xouc; A, B, T [iexpeT- 6 A apa xwv A, B, T xoivov [iexpov eaxiv. Xeyw 5r|, oxi xal A B C D E F Let A, B, and C be the three given numbers (which are) not prime to one another. So it is required to find the greatest common measure of A, B, and C. For let the greatest common measure, D, of the two (numbers) A and B have been taken [Prop. 7.2]. So D either measures, or does not measure, C. First of all, let it measure (C). And it also measures A and B. Thus, D 197 ETOIXEIfiN C- ELEMENTS BOOK 7 (ifyioTov. si yap ^Vj eoxiv 6 A xwv A, B, T ^(iyiaTov xoivov ^tETpov, ^exprjaei tic; touc; A, B, T dpid^touc; dpidfjioc; ^eic^cov wv tou A. jiexpeixw, xal eotco 6 E. etc! ouv 6 E touc; A, B, r [lexpsT, xal touc; A, B apa \iETprpe^ xal to twv A, B dpa ^liyioTov xoivov (jiETpov \xeTpf\aei. to 8s tGv A, B jieyiaTov xoivov ^iSTpov scttIv 6 A- 6 E apa tov A [iETpsT 6 ^sii^cov tov eXdaaova - ouep eotiv dSuvaTov. oux apa touc; A, B, T dpid^touc; dpi-djioc; tic; \xexpr\oei ^.aCtov wv tou A- 6 A apa twv A, B, T jieyioTov ecru xoivov ^STpov. Mf) [iSTpeiTW 8r) 6 A tov E XsyM TtpfiTov, oti oi T, A oux siai KpwToi Tipoc; dXXr|Xou<;. excel yap oi A, B, T oux eiai KptOTOi Tipoc dXXr|Xouc;, [iSTpr|aei tic; auTouc; dpi%6c;. 6 8r) touc; A, B, T ^.ETpwv xal touc; A, B |i£Tpr|aei, xal to twv A, B ^iyicrcov xoivov jieTpov tov A \iSTpr\ae\.- ^.ETpei 8s xal tov T- touc; A, T apa dpi%ouc; dpi-d^oc; tic; ^STpr]oei- oi A, r apa oux eiai TipoToi Ttpoc; dXXr|Xouc;. eiXricpiDw ouv auTOv to ^syiaTov xoivov jiSTpov 6 E. xal end 6 E tov A [LejpeT, 6 Se A touc; A, B ^ETpeT, xal 6 E apa touc; A, B ^.ETpeT - ^(.ETpei 8e xal tov E 6 E apa touc; A, B, T (iSTpeT. 6 E apa iSv A, B, T xoivov ecru (iETpov. Xeyco br\, oti xal ^eyiaTov. si yap u/] ecruv 6 E t£Sv A, B, T to ^leyiaTov xoivov [iSTpov, \ieTpr\aei tic; touc; A, B, T dpid^ouc; dpnD^toc; ^.EiCwv <2>v tou E. [iSTpeiTO, xal sgto 6 Z. xal ETtd 6 Z touc; A, B, r ^iSTpei, xal touc; A, B ^.ETpei' xal to twv A, B apa ^iiyiaTov xoivov [iSTpov \iexp'f]aei. to Se t£>v A, B ^eyiaTOv xoivov ^.ETpov sotIv 6 A- 6 Z dpa tov A \isxpev [izxpei 8s xal tov T- 6 Z apa touc; A, T ^i£Tper xal to twv A, T apa ^iiyiaTov xoivov (iSTpov \LeTpf\aei. to 8e tcjv A, T jieyiaTOv xoivov [iSTpov eotIv 6 E - 6 Z apa tov E [lexpei 6 [le'i^v tov eXdaaova- ouep eotIv dSuvaTov. oux apa touc; A, B, T dpid^touc; dpid^ioc; tic; \iexpr\oei jieiCwv wv tou E - 6 E apa tcjv A, B, r jieyiaTov eoti xoivov ^ETpov onep e8ei Seic;ai. measures A, B, and C. Thus, D is a common measure of A, B, and C. So I say that (it is) also the greatest (common measure) . For if D is not the greatest common measure of A, B, and C then some number greater than D will measure the numbers A, B, and C. Let it (so) measure {A, B, and C), and let it be E. Therefore, since E measures A, B, and C, it will thus also measure A and B. Thus, it will also measure the greatest common mea- sure of A and B [Prop. 7.2 corr.]. And D is the greatest common measure of A and B. Thus, E measures D, the greater (measuring) the lesser. The very thing is impossi- ble. Thus, some number which is greater than D cannot measure the numbers A, B, and C. Thus, D is the great- est common measure of A, B, and C. So let D not measure C. I say, first of all, that C and D are not prime to one another. For since A, B, C are not prime to one another, some number will measure them. So the (number) measuring A, B, and C will also measure A and B, and it will also measure the greatest common measure, D, of A and B [Prop. 7.2 corr.]. And it also measures C. Thus, some number will measure the numbers D and C. Thus, D and C are not prime to one another. Therefore, let their greatest common measure, E, have been taken [Prop. 7.2] . And since E measures D, and D measures A and B, E thus also measures A and B. And it also measures C. Thus, E measures A, B, and C. Thus, E is a common measure of A, B, and C. So I say that (it is) also the greatest (common measure). For if E is not the greatest common measure of A, B, and C then some number greater than E will measure the num- bers A, B, and C. Let it (so) measure (A, B, and C), and let it be F. And since F measures A, B, and C, it also measures A and B. Thus, it will also measure the great- est common measure of A and B [Prop. 7.2 corr.]. And D is the greatest common measure of A and B. Thus, F measures D. And it also measures C. Thus, F measures D and C. Thus, it will also measure the greatest com- mon measure of D and C [Prop. 7.2 corr.]. And E is the greatest common measure of D and C. Thus, F measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than E does not measure the numbers A, B, and C. Thus, E is the greatest common measure of A, B, and C. (Which is) the very thing it was required to show. "Auac; dpidjioc; navToc; dpiduou 6 eXdaawv tou ^id^ovoc; f]TOi [Jispoc; egtIv fj [L&pr\. 'EaTwaav 8uo dpi$uol oi A, BT, xal scttm eXdaawv 6 Br- Xeyw, oti 6 BT tou A j^toi ^lepoc; e<xclv fj [lept]. Proposition 4 Any number is either part or parts of any (other) num- ber, the lesser of the greater. Let A and BC be two numbers, and let BC be the lesser. I say that BC is either part or parts of A. 198 ETOIXEIfiN C- ELEMENTS BOOK 7 Oi A, Br yap fjxoi Ttpaixoi npbc, dXXrjXouc; eiaiv f] ou. eoxcooav rcpoxepov oi A, Br Ttpfixoi upog dXXiqXouc;. Biai- pei&£VToc; 8/) xou Br el? iac, ev auxfi ^tovdBa<; eoxai exdoxr) [iovac x«v sv xG Br [izpoc, xi xoO A- wax£ (iepr) eoxiv 6 Br xou A. For A and BC are either prime to one another, or not. Let A and BC, first of all, be prime to one another. So separating BC into its constituent units, each of the units in BC will be some part of A. Hence, BC is parts of A. Bi B T A Mr] eoxcooav Sr| oi A, Br upcoxoi 7ip6<; dXXrjXout;' 6 8r) Br xov A fjxoi [iztpzi fj ou (jiexpel. si [ie\ ouv 6 Br xov A ^lexpeT, [ispoz eoxiv 6 Br xoO A. ei 8e ou, eiAfjcpi&M xfiv A, Br [icyioxov xoivov [iexpov 6 A, xai 8ir]piqoiL>M 6 Br si<z xou? xG A Taoue; xoug BE, EZ, Zr. xai enei 6 A xov A (jiexpsi, [izpoc, eoxiv 6 A xou A- i'ooc; 8e 6 A exdoxip xfiv BE, EZ, Zr- xai exaoxoc; dpa xfiv BE, EZ, Zr xou A ^epoc eoxiv a>oxe [ispi] eoxiv 6 Br xou A. "Anotz dpa dpiduoc; Tiavxog api^ou 6 eXdoowv xou (jieiCovog fjxoi ^epoc; eoxiv fj piepr)- onep eBei 8el^ai. A D So let A and BC be not prime to one another. So BC either measures, or does not measure, A. Therefore, if BC measures A then BC is part of A. And if not, let the greatest common measure, D, of A and BC have been taken [Prop. 7.2], and let BC have been divided into BE, EF, and FC, equal to D. And since D measures A, D is a part of A And D is equal to each of BE, EF, and FC. Thus, BE, EF, and FC are also each part of A. Hence, BC is parts of A. Thus, any number is either part or parts of any (other) number, the lesser of the greater. (Which is) the very thing it was required to show. Proposition 5* 'Edv dpii^oc; dpi^ou ^epo<; fj, xai exepoc; exepou xo auxo \xepoc, fj, xai ouvajicpoxepoc; ouva^cpoxepou xo auxo [iepoc; eoxai, onep 6 elg xou evoc;. B H r T E z A A Apiduoc; yap 6 A [dpiduou] xou Br ^tepoc; eaxw, xai If a number is part of a number, and another (num- ber) is the same part of another, then the sum (of the leading numbers) will also be the same part of the sum (of the following numbers) that one (number) is of an- other. B G tE H A D For let a number A be part of a [number] BC, and 199 ETOIXEIfiN C- ELEMENTS BOOK 7 exepoc; 6 A exepou xou EZ to auxo ^tepoc;, ojtep 6 A xou Br- Xeyo, oxi xai auva^tcpoxepoc; 6 A, A auvajicpoxepou xou Br, EZ xo auxo \iepoc, eaxiv, ojtep 6 A xou Br. 'Etc! yap, o jiepoc; eaxlv 6 A xou Br, xo auxo ^tepoc; eaxl xai 6 A xou EZ, oaoi apa eialv ev xfii Br dpi-d^ol tool iu A, xoaouxoi eiai xai ev xQ EZ dpid^tol taoi xw A. 5ifjpr]C7d« 6 [Lev Br Etc; xouc; x£5 A taouc; xouc; BH, Hr, 6 8e EZ eic; xouc; xo A i'aouc; xouc; EO, 8Z- eaxai 8r) taov xo iiXfj'doc; xov BH, Hr xw TtXyydei xwv E6, 8Z. xai £7tei taoc; eaxlv 6 \ik\ BH xw A, 6 8e E8 xw A, xai oi BH, EG dpa xoTc; A, A tool. Sid xd auxd 8f] xai oi Hr, 0Z xoTc A, A. oaoi dpa [eialv] ev xo Br dpid^tol tool xo A, xoaouxoi rial xai ev xoTc; Br, EZ tool xou; A, A. oaaitXaaiwv apa eaxlv 6 Br xou A, xoaauxaTiXaaiov eaxl xai auvajicpoxepoc; 6 Br, EZ auva^tcpoxepou xou A, A. o apa ^tepoc; eaxlv 6 A xou Br, xo auxo [lepoc, eaxl xai auvajicpoxepoc; 6 A, A auva^icpoxepou xou Br, EZ- oTtep e8ei SeTc;ai. another (number) D (be) the same part of another (num- ber) EF that A (is) of BC. I say that the sum A, D is also the same part of the sum BC, EF that A (is) of BC. For since which(ever) part A is of BC, D is the same part of EF, thus as many numbers as are in BC equal to A, so many numbers are also in EF equal to D. Let BC have been divided into BG and GC, equal to A, and EF into EH and HF, equal to D. So the multitude of (divisions) BG, GC will be equal to the multitude of (di- visions) EH, HF. And since BG is equal to A, and EH to D, thus BG, EH (is) also equal to A, D. So, for the same (reasons), GC, HF (is) also (equal) to A, D. Thus, as many numbers as [are] in BC equal to A, so many are also in BC, EF equal to A, D. Thus, as many times as BC is (divisible) by A, so many times is the sum BC, EF also (divisible) by the sum A, D. Thus, which(ever) part A is of BC, the sum A, D is also the same part of the sum BC, EF. (Which is) the very thing it was required to show. t In modem notation, this proposition states that if a = (l/n)b and c = (1/n) d then (a + c) = (1/n) (6 + d), where all symbols denote numbers. Proposition 6 f 'Edv dpid^ioc; dpi'djiou (iepr) fj, xai exepoc; exepou xa auxd [iepr] fj, xai auvajicpoxepoc; auvapicpoxepou xa auxd [ispr] eaxai, onep 6 eic; xou evoc;. A H 1 B 1 r A iE z Apidjioc; yap 6 AB dpiiD^tou xou T [icpr] eaxw, xai exepoc; 6 AE exepou xou Z xa auxd ^epr), anep 6 AB xou T- Xeyto, oxi xai auva^xcpoxepoc; 6 AB, AE auva^xcpoxepou xou T, Z xd auxd \iipr\ eaxiv, drcep 6 AB xou T. Tkel yap, a [iepf] eaxlv 6 AB xou T, xd auxd [iepr\ xai 6 AE xou Z, oaa dpa eaxlv ev xw AB [iepf] xou T, xoaauxd eaxi xai ev xw AE [ispf] xou Z. 8ir]pr]a'f)w 6 [Lev AB eic; xd xou T \iept) xd AH, HB, 6 8e AE eic; xd xou Z [iepf] xd A0, 6E- eaxai 8r] laov xo TtXrydoc; x«v AH, HB xw txXt)tl>ei xwv A0, 6E. xai etxel, o uepoc; eaxlv 6 AH xou T, xo If a number is parts of a number, and another (num- ber) is the same parts of another, then the sum (of the leading numbers) will also be the same parts of the sum (of the following numbers) that one (number) is of an- other. A G B D H For let a number AB be parts of a number C, and an- other (number) DE (be) the same parts of another (num- ber) F that AB (is) of C. I say that the sum AB, DE is also the same parts of the sum C, F that AB (is) of C. For since which (ever) parts AB is of C, DE (is) also the same parts of F, thus as many parts of C as are in AB, so many parts of F are also in DE. Let AB have been divided into the parts of C, AG and GB, and DE into the parts of F, DH and HE. So the multitude of (divisions) AG, GB will be equal to the multitude of (divisions) DH, 200 ETOIXEIfiN C- ELEMENTS BOOK 7 auxo ^epoc; eaxl xai 6 A0 xou Z, 8 dpa \±spoz eaxlv 6 AH xoO T, to auxo \Lspoz eaxl xai auva|icp6xepo<; 6 AH, AO auva^tcpoxepou xou T, Z. Sid xa auxa 8f] xai 6 ^epo<; eaxlv 6 HB xoO r, xo auxo ^epoc eaxl xai auva^cpoxepo<; 6 HB, 9E auva^icpoxepou xou T, Z. a apa ^epr) eaxlv 6 AB xoO T, xa auxa (jiepr) eaxl xai auva^icpoxepog 6 AB, AE auva^tcpoxepou xou T, Z- ouep eBei 8el£ai. And since which (ever) part AG is of C, DH is also the same part of F, thus which(ever) part AG is of C, the sum AG, DH is also the same part of the sum C, F [Prop. 7.5]. And so, for the same (reasons), which(ever) part GB is of C, the sum GB, HE is also the same part of the sum C, F. Thus, which (ever) parts AB is of C, the sum AB, DE is also the same parts of the sum C, F. (Which is) the very thing it was required to show. t In modem notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a + c) = (m/n) (b + d), where all symbols denote numbers. 'Edv dpL-djioc; dpi-djiou [icpoc fj, onep dcpaipe-dslc; dcpai- pei^evxoc;, xai 6 Xoltt6<; xou Xoittou xo auxo [lepoc. eaxai, OTtep 6 oXoc; xou oXou. A E B i — i 1 H r Z A I 1 1 1 Apii9(ji6(; jap 6 AB dpi-fijiou xou IA |iepo<; eaxo, oitep dcpaips-dric; 6 AE dcpaipeiJevxoc; xou TZ- Xeyw, oxi xai Xomog 6 EB Xoittou xou ZA xo auxo y.epoc, eoxiv, oTisp oXoc; 6 AB oXou xou TA. "0 yap [lepoz iaxlv 6 AE xou TZ, xo auxo [izpoQ eaxco xai 6 EB xou TH. xai ine'i, o ^.epoc; saxlv 6 AE xou TZ, xo auxo \xspoz eaxl xai 6 EB xou TH, o apa ^.epo<; eaxlv 6 AE xou TZ, xo auxo ^iepo<; eaxl xai 6 AB xou HZ. o Be ^tepo<; eaxlv 6 AE xou TZ, xo auxo \±spoz (moxeixai xai 6 AB xou TA- o apa \±epoq eaxl xai 6 AB xou HZ, xo auxo \±epoq eaxl xai xou TA- l'ao<; apa eaxlv 6 HZ ifi TA. xoivo<; dtpr)pr]a , dw 6 TZ' Xoikoz apa 6 Hr Xomw xw ZA eaxiv Xaoc,. xai etxeC, o ^.epoc eaxlv 6 AE xou TZ, xo auxo (icpog [eaxl] xai 6 EB xou Hr, Xaoc, 8e 6 Hr iu ZA, S apa ^iepo<; eaxlv 6 AE xou TZ, xo auxo |iepo<; eaxl xai 6 EB xou ZA. dXXa o ^tepo<; eaxlv 6 AE xou TZ, xo auxo ^.epoc; eaxl xai 6 AB xou FA- xai Xom6<; apa 6 EB Xomou xou ZA xo auxo ^tepo<; eaxlv, oTiep oXoc, 6 AB oXou xou TA- onep e8ei SeT^ai. Proposition 7 f If a number is that part of a number that a (part) taken away (is) of a (part) taken away then the remain- der will also be the same part of the remainder that the whole (is) of the whole. A E B i — i 1 G C F D i 1 1 1 For let a number AB be that part of a number CD that a (part) taken away AE (is) of a part taken away CF. I say that the remainder EB is also the same part of the remainder FD that the whole AB (is) of the whole CD. For which(ever) part AE is of CF, let EB also be the same part of CG. And since which(ever) part AE is of CF, EB is also the same part of CG, thus which(ever) part AE is of CF, AB is also the same part of GF [Prop. 7.5]. And which(ever) part AE is of CF, AB is also assumed (to be) the same part of CD. Thus, also, which (ever) part AB is of GF, (AB) is also the same part of CD. Thus, GF is equal to CD. Let CF have been subtracted from both. Thus, the remainder GC is equal to the remainder FD. And since which (ever) part AE is of CF, EB [is] also the same part of GC, and GC (is) equal to FD, thus which(ever) part AE is of CF, EB is also the same part of FD. But, which(ever) part AE is of CF, AB is also the same part of CD. Thus, the remain- der EB is also the same part of the remainder FD that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show. t In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a — c) = (1/n) (6— d), where all symbols denote numbers. 7]'. Proposition 8 f 'Eav dpidjioc; dpidjiou \ispt) rj, drcep dcpaipeiitelc; dcpai- If a number is those parts of a number that a (part) pe$evxo<;, xai 6 Xolttoc xou Xoittou xa auxa [iepr] eaxai, taken away (is) of a (part) taken away then the remain- ditep 6 8Xo<; xou oXou. der will also be the same parts of the remainder that the 201 ETOIXEIfiN C- ELEMENTS BOOK 7 whole (is) of the whole. r Z A C F D I 1 1 I 1 1 H M K N GMKNH i 1 — i 1 — i i 1 — i 1 — i A A E B i 1 1 1 Apid^io? yap o AB dpi/d^iou tou IA \±epr\ screw, abtep acpaipsdsle; 6 AE dcpaipsiSsvToc; tou TZ- Xsyto, oxi xal Xomoc 6 EB Xoittou xou ZA Ta auTa [ispf] sot(v, arcsp oXoc; 6 AB oXou tou TA. Ksitxdo yap tG AB iaoc 6 H8, a dpa \±£pr] scrav 6 H9 tou TA, Ta ai)Ta ^ispr] sera xal 6 AE tou TZ. 8ir)pr]a , do 6 ^isv H9 si? Ta tou TA [leprj Ta HK, K9, 6 Ss AE tic, Ta tou TZ [itpr\ Ta AA, AE- saTai Sr] laov to TtXrydoc; t«v HK, K0 Tip TcXrjn&ei tcov AA, AE. xal inti, 6 ^epog scrav 6 HK tou TA, to auTo y.epoc sera xal 6 AA tou TZ, jis^tov 8s 6 TA tou TZ, [isi^tov apa xal 6 HK tou AA. xsicrdio Ttp AA faoc; 6 HM. o apa ^ispoc; saxlv 6 HK tou TA, to auTo ^ispoc ecra xal 6 HM tou TZ- xal Xoinbz apa 6 MK XomoO tou ZA to auTo ^epoc sotiv, OTtsp oXoc; 6 HK oXou tou TA. TtdXiv stcsl, o fispoc scrav 6 K9 tou TA, to auTo ^ispo<; sera xal 6 EA tou TZ, [idCwv Ss 6 TA tou TZ, ^si^cov dpa xal 6 6K tou EA. xsicrdco to EA taoc. 6 KN. o apa ^tspo<; scrav 6 K0 tou TA, to auTo ^tepo<; sera xal 6 KN tou TZ- xal Xoitcoc dpa 6 N9 Xoitiou tou ZA to auTo [izpoc, scrav, oitsp oXo<; 6 K9 oXou tou TA. sBsix'dr) 8s xal Xomo<; 6 MK Xoitiou tou ZA to ai)To ^tepot; <J>v, OTisp okoc 6 HK oXou tou TA- xal auva^tcpoTspoc; apa 6 MK, N9 tou AZ Ta auTa [itpf] scrav, djtsp okoc, 6 9H oXou tou TA. Taoc, 8s auva^tcpoTspoc [isv 6 MK, N9 tw EB, 6 Bs 9H to BA- xal Xomoc. dpa 6 EB Xoitiou tou ZA Ta auTa [itpf] scrav, djisp 6Xoc. 6 AB oXou tou IA- oitsp eBsi 8su;ai. A L E B i 1 1 1 For let a number AB be those parts of a number CD that a (part) taken away AE (is) of a (part) taken away CF. I say that the remainder EB is also the same parts of the remainder FD that the whole AB (is) of the whole CD. For let GH be laid down equal to AB. Thus, which (ever) parts GH is of CD, AE is also the same parts of CF. Let GH have been divided into the parts of CD, GK and KH, and AE into the part of CF, AL and LE. So the multitude of (divisions) GK, KH will be equal to the multitude of (divisions) AL, LE. And since which (ever) part GK is of CD, AL is also the same part of CF, and CD (is) greater than CF, GK (is) thus also greater than AL. Let GM be made equal to AL. Thus, which(ever) part GK is of CD, GM is also the same part of CF. Thus, the remainder MK is also the same part of the remainder FD that the whole GK (is) of the whole CD [Prop. 7.5]. Again, since which(ever) part KH is of CD, EL is also the same part of CF, and CD (is) greater than CF, HK (is) thus also greater than EL. Let KN be made equal to EL. Thus, which(ever) part KH (is) of CD, KN is also the same part of CF. Thus, the remain- der NH is also the same part of the remainder FD that the whole KH (is) of the whole CD [Prop. 7.5]. And the remainder MK was also shown to be the same part of the remainder FD that the whole GK (is) of the whole CD. Thus, the sum MK, NH is the same parts of DF that the whole HG (is) of the whole CD. And the sum MK, NH (is) equal to EB, and HG to BA. Thus, the remainder EB is also the same parts of the remainder FD that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show. t In modem notation, this proposition states that if a = (m/n)b and c = (m/n) d then (a - c) = (m/n) (b - d), where all symbols denote numbers. Proposition 9 1 'Edv dpi%6c dpi%ou [lepoc ?j, xal STspog STspou to If a number is part of a number, and another (num- auTo ^spoc; fj, xal svaXXdc;, o ^spoc; scrav fj (ispr) 6 np&Toc ber) is the same part of another, also, alternately, tou Tphou, to auTo jjispoc. saTai f\ Ta auTa [iepr\ xal 6 which(ever) part, or parts, the first (number) is of the BsuTspoc. tou TETdpTou. third, the second (number) will also be the same part, or 202 ETOIXEIfiN C- ELEMENTS BOOK 7 B H A A Apidjioc; yap 6 A dpidjioO xou Br \±spoz eaxo, xal exe- poc 6 A exepou xou EZ to auxo [iepo<;, ojtep 6 A xou Br^ Xeyw, oxi xai evaXXd?, o \iepoq eaxlv 6 A xou A fj ^epr), xo auxo \xepoc, eaxl xal 6 Br xou EZ f] ^epr). 'End yap o ^spo<; eaxlv 6 A xou Br, xo auxo ^.epoc eaxl xal 6 A xou EZ, oaoi apa eialv ev iw Br dpid^ol laoi x£> A, TooouTo'i eiai xal ev xw EZ Taoi xc5 A. 8iy]pfja , d« 6 [lev Br tic, xou? xc5 A laouc xou<; BH, Hr, 6 Se EZ el? xou<; iu A laouc; xou<; E6, 0Z- eaxai 8f] laov xo TiXfjiEkK xGv BH, Hr iS TiXVjflei xwv EG, ez. Kai ekei Taoi elalv oi BH, Hr apid^iol dXXfjXoic;, eial 8e xal oi EG, 0Z dpi/d^ol i'aoi dXXfjXou;, xa[ eaxiv i'aov xo TiXfj'dot; iwv BH, Hr xCS irXfj'dei xwv E9, 0Z, 8 dpa y.epoc, eaxlv 6 BH xou E9 fj [iepf], xo auxo ^tepoc; eaxl xal 6 Hr xou 9Z fj xd auxa \i£pr\ % &axe xal 6 ^iepo<; eaxlv 6 BH xou E9 fj [i£pr\, xo auxo ^tepo<; eaxl xal auva^icp6xepo<; 6 Br auva^tcpoxepou xou EZ fj xd auxa [ispr\. Xooc, Se 6 uev BH tu A, 6 8e E9 x£> A' o apa ^tepo<; eaxlv 6 A xou A fj ^epr], xo auxo ^lepoc eaxl xal 6 Br xou EZ fj xd auxa [i£pr\- onep e8ei 8el^ai. the same parts, of the fourth. B E T H A D For let a number A be part of a number BC, and an- other (number) D (be) the same part of another EF that A (is) of BC. I say that, also, alternately, which(ever) part, or parts, A is of D, BC is also the same part, or parts, of EF. For since which(ever) part A is of BC, D is also the same part of EF, thus as many numbers as are in BC equal to A, so many are also in EF equal to D. Let BC have been divided into BC and GC, equal to A, and EF into -EiJ and HF, equal to £>. So the multitude of (di- visions) BC, GC will be equal to the multitude of (divi- sions) EH, HF. And since the numbers BC and GC are equal to one another, and the numbers EH and HF are also equal to one another, and the multitude of (divisions) BG, GC is equal to the multitude of (divisions) EH, HC, thus which (ever) part, or parts, BG is of EH, GC is also the same part, or the same parts, of HF. And hence, which (ever) part, or parts, BG is of EH, the sum BC is also the same part, or the same parts, of the sum EF [Props. 7.5, 7.6]. And BG (is) equal to A, and EH to D. Thus, which(ever) part, or parts, A is of D, BC is also the same part, or the same parts, of EF. (Which is) the very thing it was required to show. t In modern notation, this proposition states that if a numbers. (1/n) b and c = (1/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote I . 'Edv dpiv}^6<; dpnf)[iou [ispf] fj, xal exepoc; exepou xd auxa [Jtepr] fj, xal evaXXd^, a \iepr\ eaxlv 6 npfixot; xou xpixou fj fjiepoc, xd auxa fiepr] eaxai xal 6 Beuxepoc xou xexdpxou fj xo auxo (iepoc- Api-djioi; yap 6 AB dpid^iou xou T \±epr) eaxco, xal exepoc; 6 AE exepou xou Z xa auxa [iepf]- Xeyw, oxi xal evaXXd^, a [iepi] eaxlv 6 AB xou AE fj ^tepo<;, xa auxa [ispf] eaxl xal 6 r xou Z fj xo auxo ^tepo<;. Proposition 10 f If a number is parts of a number, and another (num- ber) is the same parts of another, also, alternately, which (ever) parts, or part, the first (number) is of the third, the second will also be the same parts, or the same part, of the fourth. For let a number AB be parts of a number C, and another (number) DE (be) the same parts of another F. I say that, also, alternately, which (ever) parts, or part, 203 ETOIXEIfiN C- ELEMENTS BOOK 7 AB is of DE, C is also the same parts, or the same part, of F. A H B r A ® E D H 'Etc! y^Pj & t^p/] eaxlv 6 AB xou T, xd auxa [ispr\ eaxl xal 6 AE xou Z, oaa apa eaxlv ev xw AB [i£pr) xoO T, xoaauxa xal ev xw AE ^ep/] xou Z. BupfjcTdw 6 [lev AB eu; xa xou r [-tepr) xa AH, HB, 6 8e AE eu; xa xou Z (jiepr) xa A0, 9E- eaxai 8f] laov xo TtXrji!)o<; ifiv AH, HB xo TcXyydei iwv A0, 0E. xal ETtei, o ^iepo<; eaxlv 6 AH xou T, xo auxo ^iepo<; eaxl xal 6 A6 xou Z, xal evaXXd?;, o jiepo<; eaxlv 6 AH xou A8 fj ^epr], xo auxo ^iepo<; eaxl xal 6 T xou Z fj xa auxa (ieprj. 8ia xa auxa Sf] xa(, 8 ^lipoc; eaxlv 6 HB xou 6E fj [iepf] 7 xo auxo jiepo<; eaxl xal 6 T xou Z fj xa auxa ^.epr]- waxe xal [8 jiepo<; eaxlv 6 AH xou A9 fj \±spr}, xo auxo ^iepo<; eaxl xal 6 HB xou 0E fj xa auxa ^epr)- xal o apa ^tepo<; eaxlv 6 AH xou AO fj ^epr], xo auxo uepo<; eaxl xal 6 AB xou AE fj xa auxa y.ept}- dXX' 8 ^tepo<; eaxlv 6 AH xou A0 fj [iepf], xo auxo [izpoc, eBelx^/] xal 6 T xou Z fj xa auxa [ispt], xal] a [dpa] ^iepr] eaxlv 6 AB xou AE fj [iepo<;, xa auxa jiepr) eaxl xal 6 T xou Z fj xo auxo jiepoc 8icep eSei Bel^ai. t In modem notation, this proposition states that if a : numbers. A G B For since which (ever) parts AB is of C, DE is also the same parts of F, thus as many parts of C as are in AB, so many parts of F (are) also in DE. Let AB have been divided into the parts of C, AG and GB, and DE into the parts of F, DH and ifi£. So the multitude of (divisions) AG, GB will be equal to the multitude of (di- visions) DH, HE. And since which (ever) part AG is of C, DH is also the same part of F, also, alternately, which (ever) part, or parts, AG is of DH, C is also the same part, or the same parts, of F [Prop. 7.9]. And so, for the same (reasons), which (ever) part, or parts, GB is of HE, C is also the same part, or the same parts, of F [Prop. 7.9]. And so [which(ever) part, or parts, AG is of DH, GB is also the same part, or the same parts, of HE. And thus, which(ever) part, or parts, AG is of DH, AB is also the same part, or the same parts, of DE [Props. 7.5, 7.6]. But, which(ever) part, or parts, AG is of DH, C was also shown (to be) the same part, or the same parts, of F. And, thus] which (ever) parts, or part, AB is of DE, C is also the same parts, or the same part, of F. (Which is) the very thing it was required to show. (m/n)b and c = (m/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote ia . Eav fj toe; oXoc; npbz oXov, ouxcoc; dcpaipeiDelt; itpo<; dcpai- peiJevxa, xal 6 Xoittoc; upog xov Xomov eaxai, tbg oXoc; upog oXov. 'Eaxa> obc; oXoc; 6 AB npoc; oXov xov TA, ouxoc dcpai- pe$el<; 6 AE npbc, dcpaipeOevxa xov TZ- Xeyto, oxi xal Xoitcoc 6 EB Ttpoc Xomov xov ZA eaxiv, 6<; oXo<; 6 AB icpog oXov xov rA. Proposition 11 If as the whole (of a number) is to the whole (of an- other), so a (part) taken away (is) to a (part) taken away, then the remainder will also be to the remainder as the whole (is) to the whole. Let the whole AB be to the whole CD as the (part) taken away AE (is) to the (part) taken away CF. I say that the remainder EB is to the remainder FD as the whole AB (is) to the whole CD. 204 ETOIXEIfiN C- ELEMENTS BOOK 7 r A E B 1 A 'Enei ecmv &c 6 AB npoc tov TA, outcoc 6 AE npoc tov rZ, 8 apa \±spo<z ecmv 6 AB tou TA f\ \xepr\, to auxo [ispoz soil xal 6 AE tou TZ fj xa aura [isprj. xal Xomoc apa 6 EB XomoO tou ZA to auTO ^iipoc eotIv f) (iepr), dnep 6 AB tou TA. eaTiv apa (be 6 EB npoc tov ZA, outmc 6 AB npoc; tov TA- onsp s8ei Bsicai. A E B 1 D J (For) since as AB is to CD, so AE (is) to CF, thus which(ever) part, or parts, AB is of CD, AE is also the same part, or the same parts, of CF [Def. 7.20]. Thus, the remainder EB is also the same part, or parts, of the remainder FD that AB (is) of CD [Props. 7.7, 7.8]. Thus, as EB is to FD, so AB (is) to CD [Def. 7.20]. (Which is) the very thing it was required to show. t In modern notation, this proposition states that if a : b :: c : d then a : b :: a — c : b — d, where all symbols denote numbers. 'Edv Saiv onoaoiouv dpi'djiol dvdXoyov, earai (be de tGv fjyou^ievwv npoc; eva tGv snojisvwv, outmc dnavTee oi f)YOU[i£voi npoc dnavrac touc snojisvoue. Proposition 12* If any multitude whatsoever of numbers are propor- tional then as one of the leading (numbers is) to one of the following so (the sum of) all of the leading (numbers) will be to (the sum of) all of the following. A B r A 'EaT«aav onoaoiouv dpid^ioi dvdXoyov oi A, B, T, A, (be; 6 A npoc tov B, out«c 6 T npoc; tov A- Xey«, oti eotiv (be; 6 A npoc; tov B, outmc oi A, T npoc; touc B, A. 'End yap scrav (be; 6 A npoc; tov B, outmc 6 T npoc tov A, o apa ^lipoe eotIv 6 A tou B rj \±epf], to ai)TO [lepoc sojI xai 6 T tou A rj \xepr]. xal ouva[i(p6Tepoc apa 6 A, T auva^itpoTepou tou B, A to auTO [ispoe eaTiv fj Ta ai)Ta [ispf], dnep 6 A tou B. eaTiv apa (be 6 A npoc tov B, outwc oi A, T npoc touc B, A- onep e5ei 5eic;ai. A B C D Let any multitude whatsoever of numbers, A, B, C, D, be proportional, (such that) as A (is) to B, so C (is) to D. I say that as A is to B, so A, C (is) to B, D. For since as A is to B, so C (is) to D, thus which(ever) part, or parts, A is of £?, C is also the same part, or parts, of D [Def. 7.20]. Thus, the sum A, C is also the same part, or the same parts, of the sum B, D that A (is) of B [Props. 7.5, 7.6]. Thus, as A is to S, so A, C (is) to B, D [Def. 7.20]. (Which is) the very thing it was required to show. 205 ETOIXEIftN C- ELEMENTS BOOK 7 t In modern notation, this proposition states that if a : b :: c : d then a:6::a + c:6 + d, where all symbols denote numbers. 'Edv Teaaapec; dpidjiol dvdXoyov Saiv, xal evaXXdc^ dvdXoyov eaovTai. Proposition 13 f If four numbers are proportional then they will also be proportional alternately. A B r A "Eaxwaav Teaaapec; apid^ioi dvdXoyov oi A, B, T, A, 6 A Tipoc; tov B, outoc; 6 T Tipoc; tov A- Xeyw, oti xal evaXXai; dvdXoyov eaovxai, (be; 6 A Tipoc; tov T, outcoc; 6 B Tipoc; tov A. Tkel ydp eaTiv (be; 6 A Tipoc; tov B, outck 6 T Tipoc; tov A, o dpa \±spoz eaTiv 6 A tou B fj \±spr\, to auTo \±spoz tov. xal 6 T tou A fj Ta ai)Ta [ispt]. evaXXai; dpa, S jiepoe; eaTiv 6 A tou T fj [iepi], to ai)To ^tepoc; larl xal 6 B tou A fj Ta auTa nepT). eaTiv dpa q<;6 A Tipoc; tov T, outw<; 6 B Tipoc; tov A- oTiep e8el 8el5ai. t In modern notation, this proposition states that if a : b :: c : d then a : 18'. 'Eav Saiv onoaoiouv dpi-djiol xal dXXoi auToTe; I'ooi to TiXfjiEtoc; auvSuo Xa^pavojisvoL xal ev tQ auTfi Xoycp, xal 5i' I'oou ev t(0 auTW Xoyw eaovTai. A' 1 A i 1 B' 1 Ei 1 ' Z' ' 'EaTwaav oTtoaoiouv dpiiSu-ol oi A, B, T xal dXXoi auToTe; Taoi to TtXfj'doc; auvSuo Xa^tpavouevoi ev iu auT« Xoyw oi A, E, Z, &>q (iev 6 A Tipoc; tov B, outw<; 6 A Tipoc; tov E, (be 5e 6 B Tipoc; tov T, outwc; 6 E Ttp6<; tov Z- Xeyw, oti xal 8i° Taou eaTiv d;6A Tipoc; tov T, outwc; 6 A Tipoc; tov Z. 'End ydp eaTiv (be; 6 A Tipoc; tov B, outmc; 6 A Tipoc; tov E, evaXXdc; dpa eaTiv (be; 6 A Tipoc; tov A, outwc 6 B Tipoc; tov E. ndXiv, inzi eaTiv (be; 6 B npoc; tov T, outwc; 6 A B C D Let the four numbers A, B, C, and D be proportional, (such that) as A (is) to B, so C (is) to D. I say that they will also be proportional alternately, (such that) as A (is) to C, so B (is) to D. For since as A is to B, so C (is) to D, thus which(ever) part, or parts, A is of B, C is also the same part, or the same parts, of D [Def. 7.20]. Thus, alterately which (ever) part, or parts, A is of C, B is also the same part, or the same parts, of D [Props. 7.9, 7.10]. Thus, as A is to C, so B (is) to D [Def. 7.20]. (Which is) the very thing it was required to show. ::b:d, where all symbols denote numbers. Proposition 14 f If there are any multitude of numbers whatsoever, and (some) other (numbers) of equal multitude to them, (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality. A i ' D 1 B i 1 E i 1 Ci 1 F i 1 Let there be any multitude of numbers whatsoever, A, B, C, and (some) other (numbers), D, E, F, of equal multitude to them, (which are) in the same ratio taken two by two, (such that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F. I say that also, via equality as A is to C, so D (is) to F. For since as A is to B, so D (is) to E, thus, alternately, as A is to D, so B (is) to E [Prop. 7.13]. Again, since as B is to C, so E (is) to F, thus, alternately, as B is 206 ETOIXEIfiN C- ELEMENTS BOOK 7 E npbc, tov Z, evaXXdi; dpa eoxlv cbc; 6 B npbc, tov E, outcoc; to E, so C (is) to F [Prop. 7.13]. And as B (is) to E, 6 T 7ipo<; tov Z. cbc; 8e 6 B Ttpoc; tov E, outwc; 6 A Ttpoc; so A (is) to D. Thus, also, as A (is) to £>, so C (is) to F. tov A- xocl cbc; apa 6 A Ttpoc; tov A, outwc; 6 V npbc, tov Thus, alternately, as A is to C, so L> (is) to F [Prop. 7.13]. Z- evaXXdi; apa eaTiv cbc; 6 A Ttpoc; tov V, outw<; 6 A Ttpoc; (Which is) the very thing it was required to show. tov Z- onep eSei BeT^oti- t In modern notation, this proposition states that if a : b :: d : e and b : c : : e : f then a : c :: d : f, where all symbols denote numbers. IS ■ Proposition 15 'Edv [lovac; dpi , f)[i6v Tiva (jtexpf], iaaxic; Be CTepoc; dpid^ioc; dXXov Tiva dpidjiov (ieTpfj, xal evaXXdc; iaaxic; f\ ^lovac; tov Tphov dprdjiov \±£Tpr]oe\. xal 6 BeuTepoc; tov TETapTov. B H i 1 1— r — i Ah E i — K — i — A — i — Z — i Movdc; yap f] A dpi%6v Tiva tov Br ^CTperao, iaaxic; 8e STspoc; dpnf)^6<; 6 A aXXov Tiva dpidjiov tov EZ ^CTpeiTcy Xeyo, oti xal evaXXdi; iaaxic; f] A ^tovdc; tov A dpi-d^ov ^iSTpeT xal 6 Br tov EZ. 'End yap iaaxic; f) A ^lovdc; tov Br dpnf)[i6v ^teTpeT xal 6 A tov EZ, oaai apa eialv ev tw BT [iovd8ec, ToaouToi elm xal ev t« EZ dpid^iol i'aoi tG A. Birjpriadio 6 [lev Br eic; tolc, ev eauTW [lovdSac; xac, BH, H9, 8r, 6 8e EZ etc; touc; tw A laouc; touc; EK, KA, AZ. eaTai 8r) i'aov to TtXTj-doc; ifiv BH, H9, 6r tw TtXyydei twv EK, KA, AZ. xal ercel laai eialv ai BH, H9, 9T [iovd8ec dXXrjXaic;, eiai 8e xal oi EK, KA, AZ dpid^tol i'aoi dXXr]Xoi<;, xai eaTiv laov to TtXfj'doc; t«v BH, H9, 9r ^iovd8tov tG TtXfj'dei t«v EK, KA, AZ dpn9[iwv, eaTai apa cbc; f] BH jiovdc; Ttpoc; tov EK dpid^ov, outck f) H9 jiovdc; Ttpoc; tov KA dpi'dfiov xal f] 9T jiovdc; upoc; tov AZ dpid^iov. eaTai apa xal cbc; eic; twv rjyou^ievcov Ttpoc; eva iSv CTto^evwv, outwc" obtavTCc; oi fjyou^tevoi Ttpoc; dTtavTac; touc; ctio^icvouc;- eaTiv apa cbc; f) BH ^tovdc; Ttpoc; tov EK dpid^iov, outmc; 6 Br upoc; tov EZ. i'ar] Be f) BH jiovdc; Tfj A (iovdSi, 6 Be EK dpid^oc; iu A dprd^w. eaTiv apa wc; f) A [iovdc; upoc; tov A dpnf)[i6v, outwc; 6 Br upoc; tov EZ. iadxic; apa f) A [lovac, tov A dpnJ^iov jiCTpeT xai 6 Br tov EZ- oTiep e8ei 8eTc;ai. If a unit measures some number, and another num- ber measures some other number as many times, then, also, alternately, the unit will measure the third num- ber as many times as the second (number measures) the fourth. B i— G — i — H — i — C — i E i— K — i— D For let a unit A measure some number BC, and let another number D measure some other number EE as many times. I say that, also, alternately, the unit A also measures the number D as many times as BC (measures) EF. For since the unit A measures the number BC as many times as D (measures) EF, thus as many units as are in BC, so many numbers are also in EF equal to D. Let BC have been divided into its constituent units, BC, GH, and HC, and EF into the (divisions) EK, KL, and LF, equal to D. So the multitude of (units) BC, GH, HC will be equal to the multitude of (divisions) EK, KL, LF. And since the units BC, GH, and HC are equal to one another, and the numbers EK, KL, and LF are also equal to one another, and the multitude of the (units) BG, GH, HC is equal to the multitude of the numbers EK, KL, LF, thus as the unit BG (is) to the number EK, so the unit GH will be to the number KL, and the unit HC to the number LF. And thus, as one of the leading (numbers is) to one of the following, so (the sum of) all of the leading will be to (the sum of) all of the following [Prop. 7.12]. Thus, as the unit BG (is) to the number EK, so BC (is) to EF. And the unit BG (is) equal to the unit A, and the number EK to the number D. Thus, as the unit A is to the number D, so BC (is) to EF. Thus, the unit A measures the number D as many times as BC (measures) EF [Def. 7.20]. (Which is) the very thing it was required to show. t This proposition is a special case of Prop. 7.9. 207 ETOIXEIftN C- ELEMENTS BOOK 7 Edv Buo dpid^iol TtoXXarcXaaidaavxec; dXXr]Xouc; tiolGoi xivac;, oi yevojjievoi. zE, aGxcSv Taoi dXXiqXoic; eaovxai. A' ' B 1 r> ' A' 1 E' ' TEaxcoaav 5uo dpi'd^oi oi A, B, xal 6 [Lev A xov B tioX- XaTiXaaidaac; xov T tcoicixo, 6 Se B xov A TtoXXaTiXaaidaac; xov A Ttoieixav Xeyco, oxi igoc; eaxlv 6 T xfl A. Tkel yap 6 A xov B TtoXXaTiXaaidaac; xov Y TteTtoirjxev, 6 B dpa xov T jiexpeT xaxd xdc; ev xfi A jiovdBac;. ^expel 8e xal f) E ^.ovdc; xov A dpi/djiov xaxd xd<; ev auxfi [lovdBac;- iadxu; dpa f\ E fiovac; xov A dpi/d^iov (jiexpeT xal 6 B xov T. evaXXdi; dpa iadxic; f] E jiovdc; xov B dpidjiov JiexpeT xal 6 A xov r. ndXiv, ETcei 6 B xov A noXXaitXaaidaac; xov A nenoiy]xev , 6 A dpa xov A JiexpeT xaxd xdc; ev xfi> B jiovdSac;. [iexpel 8e xal i\ E fiovac; xov B xaxd xdc; ev auxcp jiovdSac;- iadxic; dpa f] E jiovdc; xov B dpid^iov JiexpeT xal 6 A xov A. iadxic; 8e f) E [iovdc; xov B dpi-d^ov ejiexpei xal 6 A xov E iadxic; dpa 6 A exdxepov xwv Y, A ^expel. iao<; dpa eaxlv 6 T xw A- oTiep eSei SeTc^ai. Proposition 16 1 " If two numbers multiplying one another make some (numbers) then the (numbers) generated from them will be equal to one another. A' ' B 1 Ci ' D 1 Ei 1 Let A and B be two numbers. And let A make C (by) multiplying B, and let B make D (by) multiplying A. I say that C is equal to D. For since A has made C (by) multiplying B, B thus measures C according to the units in A [Def. 7.15]. And the unit E also measures the number A according to the units in it. Thus, the unit E measures the number A as many times as B (measures) C. Thus, alternately, the unit E measures the number B as many times as A (mea- sures) C [Prop. 7.15]. Again, since B has made D (by) multiplying A, A thus measures D according to the units in B [Def. 7.15]. And the unit E also measures B ac- cording to the units in it. Thus, the unit E measures the number B as many times as A (measures) D. And the unit E was measuring the number B as many times as A (measures) C. Thus, A measures each of C and D an equal number of times. Thus, C is equal to D. (Which is) the very thing it was required to show. t In modern notation, this proposition states that ab = ba, where all symbols denote numbers. 'Edv dpid^oc; Buo dpid^ouc; TtoXXaTiXaaidaac; Ttoifj xivac, oi ysvo^ievoi zl, auxGv xov auxov e^ouai Xoyov xolc; iroXXa- TtXaaiaadeTaLv. A' ' b 1 r 1 A i 1 E' 1 Z — ' Apid^ioc; Y a P ° A Suo dpid^iouc; xouc; B, Y TtoXXa- TiXaaidaac; xouc; A, E Ttoieixw Xeyw, oxi eaxlv wc; 6 B Ttpoc; xov T, ouxwc; 6 A Ttpoc; xov E. Tkel yap 6 A xov B TtoXXaTiXaaidaac; xov A TteTtoirjxev, 6 B dpa xov A ^expel xaxd xd<; ev iu A ^lovdSac;. JiexpeT Proposition 17 f If a number multiplying two numbers makes some (numbers) then the (numbers) generated from them will have the same ratio as the multiplied (numbers) . A' ' B i C 1 1 D 1 E i 1 F i — i For let the number A make (the numbers) D and E (by) multiplying the two numbers B and C (respec- tively). I say that as B is to C, so D (is) to E. For since A has made D (by) multiplying B, B thus measures D according to the units in A [Def. 7.15]. And 208 ETOIXEIfiN C- ELEMENTS BOOK 7 Be xoa f\ Z ^iovd<; xov A dpii9[i6v xaxd xdc; ev auxw [lovdSac iadxic; apa i] Z jiovag xov A dpidjiov ^.expeT xal 6 B xov A. eaxiv apa cb<; f) Z uovdc Ttpoc; xov A dpidjiov, ouxco<; 6 B Ttpoc; xov A. Bid xd auxd Br] xal (bt; f] Z ^.ovdc; Ttp6<; xov A dpid^iov, ouxwc; 6 T Ttpoc; xov E- xal foz apa 6 B Ttpoc; xov A, ouxwc; 6 T Ttpoc; xov E. evaXXd^ apa eaxlv foe, 6 B Ttpoc; xov T, ouxcoc; 6 A Ttpoc; xov E- oTtep eSei Bel^ai. the unit F also measures the number A according to the units in it. Thus, the unit F measures the number A as many times as B (measures) D. Thus, as the unit F is to the number A, so B (is) to D [Dei. 7.20] . And so, for the same (reasons), as the unit F (is) to the number A, so C (is) to E. And thus, as B (is) to D, so C (is) to E. Thus, alternately, as B is to C, so D (is) to E [Prop. 7.13]. (Which is) the very thing it was required to show. t In modern notation, this proposition states that if d = a b and e = ac then d : e :: b : c, where all symbols denote numbers. IT]'. 'Edv Buo dpidjiol dpn&jjiov xiva TtoXXaTtXaaidaavxec; tiolGoi xivac;, oi yevo^ievoi ec; auxfiv xov auxov l^oum Xoyov xolc; TtoXXaTtXaaidaaaiv. A' ' B 1 r> 1 Ai 1 Ei 1 Auo yap dpid^ol oi A, B dpi'djiov xiva xov T TtoXXa- TtXaaidaavxec; xouc; A, E Ttoieixwaav Xeyio, oxi eaxiv foe, 6 A Ttpoc xov B, ouxwc; 6 A Ttpoc; xov E. 'Etcei yap 6 A xov T TtoXXaTtXaaidaac; xov A TtCTtoirjxev, xal 6 T apa xov A TtoXXaTtXaaidaac; xov A TtCTto[r)xev. Bid xd auxd Br] xal 6 T xov B TtoXXaTtXaaidaac; xov E TteTtoirjxev. dpid^ioc; Br] 6 T Buo dpid^iouc; xouc; A, B TtoXXaTtXaaidaac; xouc; A, E Tt£Tio(r]X£v. eaxiv apa foe, 6 A Ttpoc; xov B, ouxoc; 6 A Ttpoc; xov E- OTtep eBei BeT^ou. Proposition 18* If two numbers multiplying some number make some (other numbers) then the (numbers) generated from them will have the same ratio as the multiplying (num- bers). A' ' B' 1 C' ' D 1 Ei 1 For let the two numbers A and B make (the numbers) D and E (respectively, by) multiplying some number C. I say that as A is to B, so D (is) to E. For since A has made D (by) multiplying C, C has thus also made D (by) multiplying A [Prop. 7.16]. So, for the same (reasons), C has also made E (by) multiplying B. So the number C has made D and E (by) multiplying the two numbers A and B (respectively) . Thus, as A is to B, so D (is) to E [Prop. 7.17]. (Which is) the very thing it was required to show. t In modern notation, this propositions states that if a c = d and bc = e then a : b :: d : e, where all symbols denote numbers. I'd'. 'Edv xeaaapec; dpid^iol dvdXoyov Soiv, 6 ex Ttpcjxou xal xexdpxou yevojievoc; dpi/duo? i'aoc; eaxai xfi ex Beuxepou xal xpixou yevopivw dprd^iSy xal edv 6 ex Ttpwxou xal xexdpxou yevojievoc; dpi/duo? laoc, fj xw ex Beuxepou xal xpixou, oi xeaaaapec; dpiduol dvdXoyov eaovxai. "Eaxioaav xeaaapec; dpid^ol dvdXoyov oi A, B, T, A, tlx; 6 A Ttpoc; xov B, ouxwc; 6 T Ttpoc xov A, xal 6 \iev A xov A TtoXXaTtXaaidaac; xov E Ttoieixio, 6 Be B xov T TtoX- XaTtXaaidaac; xov Z Ttoieixor Xeyto, oxi'i'aoc; eaxlv 6 E xQ Z. Proposition 19* If four number are proportional then the number cre- ated from (multiplying) the first and fourth will be equal to the number created from (multiplying) the second and third. And if the number created from (multiplying) the first and fourth is equal to the (number created) from (multiplying) the second and third then the four num- bers will be proportional. Let A, B, C, and D be four proportional numbers, (such that) as A (is) to B, so C (is) to D. And let A make E (by) multiplying D, and let B make F (by) multiplying C. I say that E is equal to F. 209 ETOIXEIfiN C- ELEMENTS BOOK 7 ABTAEZH ABCDEFG 'O yap A xov E TtoXXaTtXaoidoac; xov H Ttoieixco. STtri ouv 6 A tov r TtoXXaTtXaoidoac; xov H TtSTtoirjXEv, xov 8e A TtoXXaTtXaoidoac; xov E TtSTtoirjxev, dpid^ioc; 8f) 6 A 86o dpiiSjiouc; xouc; E, A TtoXXaTtXaoidoac; xouc; H, E TteTtoirjxev. eoxiv dpa u<; 6 T Ttpoc; xov A, ouxioc; 6 H Ttpoc; xov E. dXX' 6 T Ttpoc; xov A, ouxgjc; 6 A Ttpoc; xov B- xal tbc; dpa 6 A Ttpoc; xov B, ouxcoc; 6 H Ttpoc; xov E. TtdXiv, tm\ 6 A xov r TtoXXaTtXaoidoac; xov H TteTtoiiqxsv, dXXd \xr]v xal 6 B xov T TtoXXaTtXaoidoac; xov Z Tt£Ttoir]X£v, Buo 8/) dpii9|jiol oi A, B dpi'djiov xiva xov E TtoXXaTtXaoidoavxsc; xouc; H, Z Tt£Ttoir|xaoiv. soxiv dpa d>c; 6 A Ttpoc; xov B, ouxioc; 6 H Ttpoc; xov Z. dXXd ^irjv xal wcoA Ttpoc; xov B, ouxgjc; 6 H Ttpoc; xov E' xal d>c; dpa 6 H Ttpoc; xov E, ouxioc; 6 H Ttpoc; xov Z. 6 H dpa Ttpoc; sxdxspov xSv E, Z xov auxov zyzi Xoyov I'ooc; dpa eoxiv 6 E xfi Z. 'Eoxio 8rj ndXiv I'ooc; 6 E x« Z- Xsyco, oxi saxlv 6c; 6 A Ttpoc; xov B, ouxwc; 6 E Ttpoc; xov A. Tcbv yap auxcbv xaxaaxeuaoiJevxwv, ind I'ooc; eoxlv 6 E xa> Z, eoxiv dpa cbc; 6 H Ttpoc; xov E, ouxcoc; 6 H Ttpoc; xov Z. dXX' tbc; ^i£v 6 H Ttpoc; xov E, ouxok 6 E Ttpoc; xov A, tbc; 8e 6 H Ttpoc; xov Z, ouxgjc; 6 A Ttpoc; xov B. xal tbc; dpa 6 A Ttpoc; xov B, ouxtoc; 6 E Ttpoc; xov A- onep eBsi 8eTc;ai. For let A make G (by) multiplying C. Therefore, since A has made G (by) multiplying C, and has made E (by) multiplying D, the number A has made G and by mul- tiplying the two numbers C and D (respectively). Thus, as C is to D, so G (is) to £ [Prop. 7.17]. But, as C (is) to D, so A (is) to B. Thus, also, as A (is) to _B, so G (is) to i?. Again, since A has made G (by) multiplying G, but, in fact, B has also made F (by) multiplying C, the two numbers A and £? have made G and F (respectively, by) multiplying some number C. Thus, as A is to B, so G (is) to F [Prop. 7.18]. But, also, as A (is) to B, so G (is) to E. And thus, as G (is) to E, so G (is) to F. Thus, G has the same ratio to each of E and F. Thus, E is equal to F [Prop. 5.9]. So, again, let E be equal to F. I say that as A is to -B, so G (is) to £>. For, with the same construction, since E is equal to F, thus as G is to E, so G (is) to F [Prop. 5.7]. But, as G (is) to £, so G (is) to £> [Prop. 7.17]. And as G (is) to F, so A (is) to B [Prop. 7.18]. And, thus, as A (is) to B, so G (is) to D. (Which is) the very thing it was required to show. t In modern notation, this proposition reads that if a : b :: c : d then ad = be, and vice versa, where all symbols denote numbers. X . Oi sXd/ioxoi dpidjjiol twv xov auxov Xoyov £)(6vxtov auxolc; [lexpouoi xouc; xov auxov Xoyov e)(ovxa<; iodxic; o xe ^d^tov xov ^id^ova xal 6 eXdoowv xov eXdooova. 'Eoxwoav yap eXd)(ioxoi dpidjioi xfiv xov auxov Xoyov EXovxcov xou; A, B oi EA, EZ- Xeyto, oxi iodxic; 6 EA xov A ^expEi xal 6 EZ xov B. Proposition 20 The least numbers of those (numbers) having the same ratio measure those (numbers) having the same ra- tio as them an equal number of times, the greater (mea- suring) the greater, and the lesser the lesser. For let CD and EF be the least numbers having the same ratio as A and B (respectively) . I say that CD mea- sures A the same number of times as EF (measures) B. 210 ETOIXEIfiN C- ELEMENTS BOOK 7 A B T r " r Et © H z A A B C G1 D E H-j F 'O TA ydp xou A oux eaxi [iepr). e£ yap 8uvax6v, screw xal 6 EZ apa xou B xa auxa [ispf] eaxiv, amp 6 TA xou A. oaa apa saxlv ev xio TA y.spt] xou A, xoaauxd eoxi xal sv xfi> EZ [ispf] xou B. 8i./]pr]crf>M 6 jji;v TA sic; xa xou A \iept] xa TH, HA, 6 Se EZ sic; xa xou B [iepf] xa EO, OZ- saxai 8/) taov xo TtXrydoc; xtov TH, HA xG TiAr^si xfiv E6, OZ. xal ztzei tool rioiv oi TH, HA dpnSjiol dXXiqXoic, rial 8e xal oi EO, OZ dpi-Ojioi Taoi dXXr|Xoic;, xai saxiv laov xo TtXrji9oc; iwv TH, HA ifi TiX^ei xtov EO, 6Z, eaxiv apa ibc; 6 TH Ttpoc; xov EO, ouxtoc; 6 HA rcpoc; xov OZ. saxai apa xal ok ric; xtov fjyoujjivtov 7tp6<; eva xtov eno^evwv, ouxtoc; anavxec; oi f|you^ievoi upoc; anavxag xouc; sttojisvouc;. eaxiv apa tog 6 TH Ttpoc; xov EO, ouxtog 6 TA Ttpoc; xov EZ' oi TH, EO apa xolc; TA, EZ ev xtp auxto Xoytp eialv eXdaaovec; ovxec; auifiv ouep eaxlv dSuvaxov uicoxeivxai yap oi FA, EZ eXd^iaxoi xtov xov auxov Xoyov e)(6vxtov auxolc;. oux apa |iipr] saxlv 6 TA xou A- jiepoc; apa. xal 6 EZ xou B xo auxo ^tepoc; eaxiv, onep 6 TA xou A- iadxic; apa 6 TA xov A ^expeT xai 6 EZ xov B- ouep e8ei 8el£ai. For CD is not parts of A. For, if possible, let it be (parts of A) . Thus, EF is also the same parts of B that CD (is) of A [Def. 7.20, Prop. 7.13]. Thus, as many parts of A as are in CD, so many parts of B are also in EF. Let CD have been divided into the parts of A, CG and CD, and EF into the parts of B, EH and HF. So the multi- tude of (divisions) CG, GD will be equal to the multitude of (divisions) EH, HF. And since the numbers CG and GD are equal to one another, and the numbers EH and HF are also equal to one another, and the multitude of (divisions) CG, GD is equal to the multitude of (divi- sions) EH, HF, thus as CG is to EH, so GD (is) to HF. Thus, as one of the leading (numbers is) to one of the following, so will (the sum of) all of the leading (num- bers) be to (the sum of) all of the following [Prop. 7.12]. Thus, as CG is to EH, so CD (is) to EF. Thus, CG and EH are in the same ratio as CD and EF, being less than them. The very thing is impossible. For CD and EF were assumed (to be) the least of those (numbers) having the same ratio as them. Thus, CD is not parts of A. Thus, (it is) a part (of A) [Prop. 7.4]. And EF is the same part of B that CD (is) oi A [Def. 7.20, Prop 7.13]. Thus, CD measures A the same number of times that EF (measures) B. (Which is) the very thing it was required to show. xa'. Oi upCSxoi upog dXXfjXouc; dpid^toi eXd/iaxoi eiai xtov xov auxov Xoyov c)(6vxtov auxoTc 'Eaxtoaav Tcptoxoi Txpoc dXXr|Xou<; apid^ol oi A, B- Xeyto, oxi oi A, B eXd)(Laxoi rim xtov xov auxov Xoyov e^ovxtov auxolc;. E£ yap \xr\, eaovxai xivec; xfiv A, B eXdaaovec; dpidjjiol ev xto auxto Xoyto ovxec; xou; A, B. eaxtoaav oi T, A. Proposition 21 Numbers prime to one another are the least of those (numbers) having the same ratio as them. Let A and B be numbers prime to one another. I say that A and B are the least of those (numbers) having the same ratio as them. For if not then there will be some numbers less than A and B which are in the same ratio as A and B. Let them be C and D. 211 ETOIXEIfiN C- ELEMENTS BOOK 7 A B r A E ABCDE Tkel ouv oi eXd/ioToi dpidjiol tov tov auxov Xoyov £)(6vtov (leTpouai xou<; tov auxov Xoyov e^ovxag iadxic; 6 xe ^.ei^ov tov jiei^ova xal 6 eXaTTOv tov cXaxxova, TouTeaxiv o ts ?]you^evo<; xov f)you^evov xal 6 en6[ie\oc, tov eTto^ievov, iadxu; apa 6 T tov A jieTpel xal 6 A tov B. oadxic 5r) 6 T tov A jieTpeT, ToaaDTai ^iovd8e<; eaToaav ev iw E. xal 6 A apa tov B \xejpel xaxa tote, ev to E ^iovd8ac;. xal E7cd 6 r tov A (iCTpel xaTa tuq ev iu E ^tovd8a<;, xat 6 E apa tov A \Lexpei xaTa Tag ev to T jiovdBac;. 8id Ta auTa 8r] 6 E xal tov B ^eTpel xaTa Tag ev iS A ^lovdBag. 6 E apa tou<; A, B (iCTpel 7tpoTou<; ovTac; Tipog dXXr)Xou<;' OTiep sotIv dSuvaTov. oux apa eaovTai Tiveg tov A, B eXdaaove<; dpiiL>[iol ev to ai)TO Xoyo ovtcc; toTc A, B. oi A, B apa eXd)(ioTo[ eloi tov tov auTov Xoyov e^ovTCov auTolc;- ojtep e8ei 8eTc;ai. x(3'. Oi sXdxiaToi dpi%ol iwv tov auTov Xoyov e^ovTOv auTolc; upoToi npbc, dXXf|Xouc; eiaiv. A' 1 B 1 r 1 Ai 1 Ei 1 TEaToaav eXdxicrcoi dprd^oi tov tov auTov Xoyov cXovtov auTou; oi A, B- Xeyo, oti oi A, B TtpoToi 7tp6<; dXXrjXoug rialv. Ei yap \vf\ eiai TtpoToi izpbz dXXiqXouc;, \LSTpr\os\. uc, auToug dprdjioc;. [leTperco, xal eaTO 6 T. xal oadxic [iev 6 T tov A [jiETpEi, ToaaDTai jiovdBec; eaToaav ev to A, Therefore, since the least numbers of those (num- bers) having the same ratio measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser — that is to say, the leading (measuring) the leading, and the following the following — C thus mea- sures A the same number of times that D (measures) B [Prop. 7.20] . So as many times as C measures A, so many units let there be in E. Thus, D also measures B accord- ing to the units in E. And since C measures A according to the units in E, E thus also measures A according to the units in C [Prop. 7.16]. So, for the same (reasons), E also measures B according to the units in D [Prop. 7.16]. Thus, E measures A and B, which are prime to one an- other. The very thing is impossible. Thus, there cannot be any numbers less than A and B which are in the same ratio as A and B. Thus, A and B are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show. Proposition 22 The least numbers of those (numbers) having the same ratio as them are prime to one another. A' 1 Bi 1 Ci 1 Di 1 Ei 1 Let A and B be the least numbers of those (numbers) having the same ratio as them. I say that A and B are prime to one another. For if they are not prime to one another then some number will measure them. Let it (so measure them), and let it be C. And as many times as C measures A, so 212 ETOIXEIfiN C- ELEMENTS BOOK 7 oadxic; 8e 6 T tov B [izxpei, xooauxai ^iovd8ec; eaxwaav ev t5 E. 'ETtri 6 T tov A [iZTpei xaxd tolc, ev ifi A (iovd8ac;, 6 r dpa tov A TtoXXaTtXaaidaac; tov A TtCTtoiiqxev. Sid toc auxd 8r) xal 6 T tov E TtoXXaTtXaaidaac; tov B TtCTtoirjxev. dpid^toc; Br] 6 T Suo dpid^touc; touc; A, E TtoXXaTtXaaidaac; touc; A, B TtCTtoirjxev eaTiv dpa tlx; 6 A Ttpoc; tov E, outoc; 6 A Ttpoc; tov B- oi A, E dpa toTc A, B ev t« auTO Xoyw siaiv eXdaaovsc; ovtsc; auTfiv oitep sotiv dBuvaTov. oux dpa touc; A, B dpidjiouc; dpi-d^oc; tic; [LSTpr]aei. oi A, B dpa TtpfiToi Ttpoc; dXXrjXouc; eioiv oTtep sSei 8eic;ai. xy'. °Edv 80o apidfjiol TtpCnoi Ttpoc; dXXr]Xouc; SSmv, 6 tov eva auTGv (jiETpGv apiS^cx; Ttpoc; tov Xoittov TtpfiToc; eaxai. A B r A 'Eaxcoaav 8uo api'djj.oi TtpcSxoi Ttpoc; aXkr\kovc oi A, B, xov Se A ^xexpsixo tic dpi%6<; 6 r- Xeyw, oxi xal oi T, B Ttp&TOi Ttpoc dXXf]Xou<; daiv. Ei yap \ir\ eiaiv oi T, B TtpcoToi Ttpoc; dXXrjXouc;, [iETpiqaei [tic] touc r, B dpiduoc;. ^tSTpeiTW, xal sotw 6 A. sitei 6 A tov T usTpei, 6 8e T tov A ^expei, xal 6 A dpa tov A ^STpei. [i€Tpei 8s xal tov B- 6 A dpa touc; A, B ^expei TtpcVtouc; ovTac; Ttpoc; aXkr\kovc onep eaTiv dSuvaTov. oux dpa touc T, B dpi'O^iouc; dpi-duoc; tic; ^exprpei. oi T, B dpa TtpCnoi Ttpoc; dXXf]Xouc; eiaiv oitep e8ei 8eTc;ai. x5'. 'Edv Suo dpi'd^ioi Ttpoc; Tiva dpidjiov itpwToi Saiv, xal 6 zi auT«v yzv6\xzvoc Ttpoc; tov auTov itpwToc; saTai. many units let there be in D. And as many times as C measures B, so many units let there be in E. Since C measures A according to the units in D, C has thus made A (by) multiplying D [Def. 7.15]. So, for the same (reasons), C has also made B (by) multiplying E. So the number C has made A and B (by) multiplying the two numbers D and E (respectively). Thus, as D is to E, so A (is) to B [Prop. 7.17]. Thus, D and E are in the same ratio as A and B, being less than them. The very thing is impossible. Thus, some number does not measure the numbers A and B. Thus, A and B are prime to one another. (Which is) the very thing it was required to show. Proposition 23 If two numbers are prime to one another then a num- ber measuring one of them will be prime to the remaining (one) . A B C D Let A and B be two numbers (which are) prime to one another, and let some number C measure A. I say that C and B are also prime to one another. For if C and B are not prime to one another then [some] number will measure C and B. Let it (so) mea- sure (them), and let it be D. Since D measures C, and C measures A, D thus also measures A. And (D) also mea- sures B. Thus, D measures A and B, which are prime to one another. The very thing is impossible. Thus, some number does not measure the numbers C and B. Thus, C and B are prime to one another. (Which is) the very thing it was required to show. Proposition 24 If two numbers are prime to some number then the number created from (multiplying) the former (two num- bers) will also be prime to the latter (number) . 213 ETOIXEIfiN C- ELEMENTS BOOK 7 A B r A E Z ABCDEF Auo ydp dpid^oi oi A, B upog xiva dpidjiov xov T np&xoi eaxtoaav, xcd 6 A xov B iraXXaTtXaaidaac; xov A ttoiclxw Xeyw, oxi oi T, A Ttpfixoi Tipog dXXr|Xou<; eiaiv. Ei yap (ir| eiaiv oi T, A upSnoi upoc; dXXr|Xouc;, [iexpiqae!. [tic;] xoug r, A dpii}|j.6<;. ^expeixw, xod eaxto 6 E. xai enei oi T, A TCpcdxoi Tipoc; dXXr|Xou<; eiaiv, xov 8s T jiexpeT xic; dpii9(ji6c 6 E, oi A, E dpa npSxoi Ttpoc; (xXkr\kovc, eiaiv. bouxiq 8/] 6 E xov A [iexpeT, xoaauxai ^lovdSeg eaxtoaav ev xw Z- xai 6 Z dpa xov A (lexpEi xaxa xdg ev xto E ^iovd5a<;. 6 E dpa xov Z TtoXXaTtXaaidaat; xov A TteTtoirjxev. dXXd (iT]v xai 6 A xov B TioXXanXaaidaag xov A Tie7io[r)X£v lao<z dpa eaxiv 6 ex xfiv E, Z xfi ex xwv A, B. edv 8s 6 utto x£5v dxpwv i'oog fj xa> Otco xcov [leawv, oi xeaaapec; dpii&jjioi dvdXoyov eiaiv eaxiv dpa cl>g 6 E Ttpoc; xov A, ouxcoc 6 B Tipoc; xov Z. oi 5e A, E upcoxoi, oi 8e npfixoi xai eXdxiaxoi, oi 8e eXdxiaxoi dpi'djioi xaiv xov auxov Xoyov e)(6vxcov auxou; (icxpouai xoug xov auxov Xoyov e^ovxac; iadxic; o xe [ieii^iov xov ^ei^ova xai 6 eXdaatov xov eXdaaova, xouxeaxiv 6 xe f)you[ievo<; xov fjyoujievov xai 6 CTrajievoc; xov eirajievov 6 E dpa xov B ^expeT. [icxpeT 8e xai xov T- 6 E dpa xoug B, T |iexpei upcoxoui; ovxac rcpoc; dXXrjXouc;- orcep eoxiv d8uvaxov. oux dpa xoug r, A dpidjiouc; dpid^toc xi<; [lexp^aei. oi T, A dpa Tipfixoi rcpoc; dXXiqXouc; eiaiv onep e8ei SeT^ai. For let A and B be two numbers (which are both) prime to some number C. And let A make D (by) multi- plying B. I say that C and D are prime to one another. For if C and D are not prime to one another then [some] number will measure C and D. Let it (so) mea- sure them, and let it be E. And since C and A are prime to one another, and some number E measures C, A and E are thus prime to one another [Prop. 7.23]. So as many times as E measures D, so many units let there be in F. Thus, F also measures D according to the units in E [Prop. 7.16]. Thus, E has made D (by) multiply- ing F [Def. 7.15]. But, in fact, A has also made D (by) multiplying B. Thus, the (number created) from (multi- plying) E and F is equal to the (number created) from (multiplying) A and B. And if the (rectangle contained) by the (two) outermost is equal to the (rectangle con- tained) by the middle (two) then the four numbers are proportional [Prop. 6.15]. Thus, as E is to A, so B (is) to F. And A and E (are) prime (to one another). And (numbers) prime (to one another) are also the least (of those numbers having the same ratio) [Prop. 7.21]. And the least numbers of those (numbers) having the same ratio measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser — that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures B. And it also measures C. Thus, E measures B and C, which are prime to one another. The very thing is impossible. Thus, some number cannot measure the numbers C and D. Thus, C and D are prime to one another. (Which is) the very thing it was required to show. xe'. 'Edv 86o dpi%oi TtpGxoi 7tp6<; dXXr|Xou<; waiv, 6 ex xou evo<; auxfiv yevojjievoc; Ttpoc; xov Xoitiov TtpGxoc; eaxai. TEaxcoaav 8uo dpi%oi Ttpwxoi npoc, dXXr|Xou<; oi A, B, xai 6 A eauxov TtoXXaTtXaaidaag xov T noieixw Xey«, oxi Proposition 25 If two numbers are prime to one another then the number created from (squaring) one of them will be prime to the remaining (number). Let A and B be two numbers (which are) prime to 214 ETOIXEIfiN C- oi B, T itpwxoi icpoc; dXX/jXouc; eiaiv. A B r A Ksicrdco ydp x£> A Taoc; 6 A. eicei oi A, B icpaixoi icpoc; dXXf]Xou<; Eiaiv, laoc, 8e 6 A t« A, xai oi A, B dpa icpGxoi icpoc; &XXr|Xou<; eiaiv exdxepoc; dpa xt3v A, A icpoc; xov B icpwxoc; eaxiv xal 6 ex xcov A, A dpa yevojievoc; icpoc; xov B icpwxoc; eaxai. 6 8e ex xov A, A yevojievoc; apidfjioc; eaxiv 6 T. oi T, B dpa icpoxoi icpoc; dXXr|Xouc; eiaiv oicep eSei SeTc^ai. 'Edv 8uo dpid^oi icpoc; 8uo dpid^iouc; d[icpoxepoi icpoc; exdxepov icpwxoi Saiv, xai oi eZ, auxwv yevojievoi icpcoxoi icpoc; dXXr]XoU(; eaovxai. A' 1 r> ' B' 1 A i ' E' ' Z' 1 Auo yap dpidjioi oi A, B icpoc; 80o dpi'djioug xouc T, A d^tcpoxepoi Ttpoc; exdxepov icpGxoi eaxcoaav, xai 6 jiev A xov B icoXXaicXaaidaac; xov E icoieixco, 6 8e T xov A icoXXaicXaaidaac; xov Z icoieixw Xeyw, oxi oi E, Z icpfixoi icpoc; dXXr]Xouc; eiaiv. 'Etc! yap exdxepoc; xov A, B icpoc xov T icpCSxoc; eaxiv, xai 6 ex x«v A, B dpa yevo^ievoc; icpoc; xov T icpwxoc; eaxai. 6 8e ex xwv A, B yevo^ievoc; eaxiv 6 E - oi E, T dpa icpfixoi icpoc; dXXf]Xou<; eiaiv. Sid xa auxa 8r) xai oi E, A icpcoxoi icpoc; dXXr]Xouc; eiaiv. exdxepoc; dpa xwv T, A Ttpoc; xov E icpwxoc; eaxiv. xai 6 ex xGv T, A dpa yevo^evoc; icpoc; xov E icpfixoc; eaxai. 6 8e ex xwv T, A yevo^tevoc; eaxiv 6 Z. oi E, Z dpa icpwxoi icpoc; dXXr]Xou<; eiaiv oicep eSei Selc^ai. ELEMENTS BOOK 7 one another. And let A make C (by) multiplying itself. I say that B and C are prime to one another. A B C D For let D be made equal to A. Since A and B are prime to one another, and A (is) equal to D, D and B are thus also prime to one another. Thus, D and A are each prime to B. Thus, the (number) created from (multily- ing) D and A will also be prime to B [Prop. 7.24]. And C is the number created from (multiplying) D and A. Thus, C and B are prime to one another. (Which is) the very thing it was required to show. Proposition 26 If two numbers are both prime to each of two numbers then the (numbers) created from (multiplying) them will also be prime to one another. A i ' C' ' B i 1 Di 1 Ei ' F ' ' For let two numbers, A and B, both be prime to each of two numbers, C and D. And let A make E (by) mul- tiplying B, and let C make F (by) multiplying D. I say that E and F are prime to one another. For since A and B are each prime to C, the (num- ber) created from (multiplying) A and B will thus also be prime to C [Prop. 7.24]. And E is the (number) cre- ated from (multiplying) A and B. Thus, E and C are prime to one another. So, for the same (reasons), E and D are also prime to one another. Thus, C and D are each prime to E. Thus, the (number) created from (multiply- ing) C and D will also be prime to E [Prop. 7.24]. And F is the (number) created from (multiplying) C and D. Thus, E and F are prime to one another. (Which is) the very thing it was required to show. 215 ETOIXEIfiN C- ELEMENTS BOOK 7 'Eav Buo dpid^ioi Tipfixoi npog dXXf|Xouc; Saiv, xal tioX- XomAaai&aai; exdxepo^ eauTov uoif) xiva, oi yevo^ievoi ec; auT<3v Tipoxoi npog dXXr]Xou<; eaovxai, xav oi e<; apxrjg tou<; yevo^tevoui; TToXXaTtXaaidaavTec; noi&ai Tivag, xdxeTvoi TipwToi Kpo<; dXXr]Xou<; eaovTai [xal del nepl touc, axpouc, touto au^paivei]. A B T A E Z Proposition 27 f If two numbers are prime to one another and each makes some (number by) multiplying itself then the num- bers created from them will be prime to one another, and if the original (numbers) make some (more numbers by) multiplying the created (numbers) then these will also be prime to one another [and this always happens with the extremes] . A B C D E F "EaTwaav Buo dpi%ol Kpcoxoi 7tp6<; dXXr|Xou<; oi A, B, xal 6 A eauTov [lev 7ioXXa7iXaaidaa<; tov T itoierao, tov Be r 7toXXaTtXaaidaa<; tov A tcoicitco, 6 Be: B eauTov [iev TtoXXanXaaidaat; tov E koieltw, tov Be E noXXaTtXaaidaac; tov Z tioisitw Xeyto, oti oi' ts T, E xal oi A, Z TtpCrcoi npbc, dXXf]Xou<; eiaiv. 'End yap oi A, B TtpCrcoi npb<z dXXr|Xou<; eiaiv, xal 6 A eauTov TioXXanXaoidoag tov T TteTto[r)xev, oi T, B apa TtpCrcoi Ttpoc; dXXr|Xou<; eiaiv. kizzi ouv oi T, B TtpcoToi Ttpoc; dXXr]Xouc; eiaiv, xal 6 B eauTov TtoXXarcXaaidaac; tov E TteTtoirjxev, oi T, E apa TtpCkoi Ttpoc; dXXiqXouc; eiaiv. TtdXiv, end oi A, B TtpfiToi Ttpoc; dXXr|Xouc; eiaiv, xal 6 B eauTov TtoXXaTtXaaidaac; tov E TteTtoirjxev, oi A, E apa jipwToi Ttpoc; dXXiqXoug eiaiv. enel ouv Buo dpiOjiol oi A, T Ttpoc; Buo dpii9[jioU(; touc; B, E djicpoTepoi Ttpoc; exaTepov TtpCrcoi eiaiv, xal 6 ex tC5v A, T apa yevojievoc; Ttpoc; tov ex tGv B, E upwToc; eoTiv. xai cotiv 6 |iev ex iSv A, T 6 A, 6 Be ex tCSv B, E 6 Z. oi A, Z apa TtpCrcoi Ttpoc; dXXrjXouc; eiaiv oTtep eBei BeT^ai. Let A and B be two numbers prime to one another, and let A make C (by) multiplying itself, and let it make D (by) multiplying C. And let B make E (by) multiplying itself, and let it make F by multiplying E. I say that C and E, and D and F, are prime to one another. For since A and B are prime to one another, and A has made C (by) multiplying itself, C and B are thus prime to one another [Prop. 7.25]. Therefore, since C and B are prime to one another, and B has made E (by) mul- tiplying itself, C and E are thus prime to one another [Prop. 7.25]. Again, since A and B are prime to one an- other, and B has made E (by) multiplying itself, A and E are thus prime to one another [Prop. 7.25]. Therefore, since the two numbers A and C are both prime to each of the two numbers B and E, the (number) created from (multiplying) A and C is thus prime to the (number cre- ated) from (multiplying) B and E [Prop. 7.26] . And D is the (number created) from (multiplying) A and C, and F the (number created) from (multiplying) B and E. Thus, D and F are prime to one another. (Which is) the very thing it was required to show. t In modern notation, this proposition states that if a is prime to b, then a 2 is also prime to b 2 , as well as a 3 to 6 3 , etc., where all symbols denote numbers. XT]'. 'Eav Buo dpi%oi TtpcoToi Ttpoc; dXXiqXouc; Saiv, xal au- vajicpoTepoc; Ttpoc; exaTepov auTCSv TtpSTOc; eaTai' xal eav auvajicpoTepoc; Ttpoc; eva Tiva auTWv TtpCrcoc; fj, xal oi zE, dpxfjc; dpi'd^ioi TtpCrcoi Ttpoc; dXXr|Xou<; eaovTai. Proposition 28 If two numbers are prime to one another then their sum will also be prime to each of them. And if the sum (of two numbers) is prime to any one of them then the original numbers will also be prime to one another. 216 ETOIXEIfiN C- ELEMENTS BOOK 7 A Br i 1 1 A' 1 Euyxeicrdoaav yap 860 dpi/d^ioi Txpoxoi npbc, dXXiqX- ouc ol AB, BP Xeyw, oxi xal auva^tcpoxepoc 6 Ar Ttpoc exdxepov xfiv AB, Br Ttpwxoc saxiv. El yap \ir\ sicuv ol TA, AB Ttpwxoi Ttpoc dXXr]Xouc, ^t£Tpf]a£i xic xouc TA, AB dpidjioc. ^.expeixw, xal saxw 6 A. ETte! ouv 6 A xouc TA, AB ^isxpeT, xal Xomov dpa xov Br ^Exprpa. [lezpei Be xal xov BA- 6 A dpa xouc AB, Br ^ie- xpel icpcoxouc ovxac icpoc dXXrjXouc 6%ep eaxlv dBuvaxov. oux dpa xouc FA, AB dpidjiouc dpid^ioc xic ^exp^aei- oi TA, AB dpa icpoxoi Ttpoc dXXr|Xouc eia'w. 8id xa auxd Br| xal oi Ar, TB Ttpwxoi npbc, dXXf|Xouc eiaiv. 6 TA dpa Ttpoc exdxepov xfiv AB, Br TtpCSxoc eaxiv. "Eaxcoaav Br] TtdXiv oi FA, AB Ttpcoxoi Ttpoc dXXr)Xouc Xeyw, 6x1 xal oi AB, Br Ttpfixoi Ttpoc dXXr]Xouc eiaiv. EE yap (Jir) eiaiv oi AB, Br Ttpwxoi npbc, dXXr]Xouc, [Lejpfiaei xic xouc AB, Br dpid^toc. ^lexpeixw, xal eaxw 6 A. xal £7i si 6 A exdxepov xwv AB, Br ^texpeT, xal oXov dpa xov TA (iexpr]oei. ^lexpeT Be xal xov AB- 6 A dpa xouc FA, AB ^.expel Ttpwxouc ovxac Ttpoc dXXf|Xouc" oTtep eaxiv dBuvaxov. oux dpa xouc AB, Br dpidjiouc dpnf)|i6c tic; ^texpr]aei. oi AB, Br dpa Ttpoxoi npbc, dXXr]Xouc eiaiv OTtep eBei SeT^ai. "Auac npCSxoc dpnJ^oc Ttpoc ditavxa dpii^ov, ov \j.t) \±e- xpeT, Kpwxoc eaxiv. A' ' Bi 1 ri 1 'Eaxw Ttpwxoc dprd^toc 6 A xal xov B [!/] jiexpeixGr Xeyio, oxi oi B, A TtpCkoi Ttpoc dXXrjXouc eiaiv. Ei yap ^ir] eiaiv oi B, A Ttpaixoi Ttpoc dXXrjXouc, ^texpf|aei tic auxouc dpid^ioc. (iexpeixw 6 T. inei 6 T xov B [icxpel, 6 Be A xov B ou ^expeT, 6 T dpa xw A oux eaxiv 6 auxoc. xal ETiei 6 T xouc B, A ^.expeT, xal xov A dpa (lexpel TtpGxov ovxa [ir] ov auxfi 6 auxoc OTtep eaxiv dBuvaxov. oux dpa xouc B, A jiexprjoei xic dpid^ioc. oi A, B dpa upwxoi npbc, dXXf]Xouc eioiv OTtep eBei SeT^ai. A B C 1 1 1 Di 1 For let the two numbers, AB and BC, (which are) prime to one another, be laid down together. I say that their sum AC is also prime to each of AB and BC. For if CA and AB are not prime to one another then some number will measure CA and AB. Let it (so) mea- sure (them), and let it be D. Therefore, since D measures CA and AB, it will thus also measure the remainder BC. And it also measures BA. Thus, D measures AB and BC, which are prime to one another. The very thing is impossible. Thus, some number cannot measure (both) the numbers CA and AB. Thus, CA and AB are prime to one another. So, for the same (reasons), AC and CB are also prime to one another. Thus, CA is prime to each of AB and BC. So, again, let CA and AB be prime to one another. I say that AB and BC are also prime to one another. For if AB and BC are not prime to one another then some number will measure AB and BC. Let it (so) mea- sure (them), and let it be D. And since D measures each of AB and BC, it will thus also measure the whole of CA. And it also measures AB. Thus, D measures CA and AB, which are prime to one another. The very thing is impossible. Thus, some number cannot measure (both) the numbers AB and BC. Thus, AB and BC are prime to one another. (Which is) the very thing it was required to show. Proposition 29 Every prime number is prime to every number which it does not measure. A' 1 B' 1 C' ' Let A be a prime number, and let it not measure B. I say that B and A are prime to one another. For if B and A are not prime to one another then some number will measure them. Let C measure (them) . Since C measures B, and A does not measure B, C is thus not the same as A. And since C measures B and A, it thus also measures A, which is prime, (despite) not being the same as it. The very thing is impossible. Thus, some number cannot measure (both) B and A. Thus, A and B are prime to one another. (Which is) the very thing it was required to 217 ETOIXEIfiN C- ELEMENTS BOOK 7 show. X'. 'Edv 860 dpid^iol TtoXXauXaaidaavTec; dXXr|Xou<; TtoiCkri xiva, tov 8e ysvojisvov z\ auxCSv ^lexprj tic; Tipfixoc; dpi%6c;, xal sva xGv z\ «PX*K (jieTpiqoeL. A' 1 B 1 r> 1 Ai 1 Ei 1 Auo yap dpi-djjiol oi A, B noXXaitXaaidaavTec; dXXr|Xouc; xov T Ttoisraoaav, tov 8e T ^leTpeiTW tic; TtpCkoc; dpi%6c; 6 A - Xfyco, oti 6 A sva iSv A, B [lETpsI. Tov yap A \±r\ (iSTpsiTW xai soti up«Toc; 6 A- oi A, A apa 7ip«Toi Ttpoc; dXXr|Xouc; riaiv. xal oadxig 6 A tov T (iSTpsI, ToaaOTai [iovdSec; eoT«oav sv tco E. end ouv 6 A tov r [LsxpeX xoltol Tag ev tG E jiovdBac;, 6 A apa tov E TioXXaTiXaaidoac; tov T TtSTtoi/jxev. dXXa [iy]\i xal 6 A tov B TioXXaTiXaoidaac; tov T Tieuoirjxsv I'ooc; apa sotIv 6 £x twv A, E tu ex twv A, B. eaTiv apa 6 A Ttpoc; tov A, outwc; 6 B Ttpoc; tov E. oi Be A, A rcpfinoi, oi 5e upwToi xal eXd)(ioToi, oi 8e eXd)(iaToi ^iSTpouai touc; tov auTov Xoyov EXovTac; iadxic; 6 ts ^idCtov tov ^.ei^ova xal 6 eXdoawv tov eXdaaova, toutsotiv o te fjyoupievoc; tov fjyoupievov xal 6 STio^ievoc; tov enojievov 6 A apa tov B (leTpsi. ojioicoc; Br| 8dc;o^ev, oti xal iav tov B \ir] (jLexpfj, tov A ^STprjaa. 6 A apa eva iSv A, B ^isTpeT- oitsp sSei Bel^ai. Proposition 30 If two numbers make some (number by) multiplying one another, and some prime number measures the num- ber (so) created from them, then it will also measure one of the original (numbers). A' 1 Bi 1 Ci 1 Di 1 Ei 1 For let two numbers A and B make C (by) multiplying one another, and let some prime number D measure C. I say that D measures one of A and B. For let it not measure A. And since D is prime, A and D are thus prime to one another [Prop. 7.29]. And as many times as D measures C, so many units let there be in E. Therefore, since D measures C according to the units E, D has thus made C (by) multiplying E [Def. 7.15]. But, in fact, A has also made C (by) multi- plying B. Thus, the (number created) from (multiplying) D and E is equal to the (number created) from (mul- tiplying) A and B. Thus, as D is to A, so B (is) to E [Prop. 7.19]. And D and A (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio) [Prop. 7.21], and the least (numbers) measure those (numbers) hav- ing the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser — that is to say, the leading (measuring) the lead- ing, and the following the following [Prop. 7.20]. Thus, D measures B. So, similarly, we can also show that if (D) does not measure B then it will measure A. Thus, D measures one of A and B. (Which is) the very thing it was required to show. Xa'. "Auac; auvdevToc; dpid[i6c; Otto npwTou tivoc; dpi'dpiou ^s- TpeiTai. 'Ecnxo auvdsvToc; dpi%6<; 6 A- Xeyw, oti 6 A (mo TipwTou tivoc; dpii9(jio0 (jiETpslTai. Tkel yap auvdsToc; eaTiv 6 A, \LZTpr\az\. tic; auTov Proposition 31 Every composite number is measured by some prime number. Let A be a composite number. I say that A is measured by some prime number. For since A is composite, some number will measure it. Let it (so) measure (A), and let it be B. And if B 218 ETOIXEIfiN C- ELEMENTS BOOK 7 6Lpv&\i6c,. ^ETpeiTW, xal eaxw 6 B. xal si \±ev np&xdz saxiv 6 B, yEyovoc; av sTr) xo £7uxax$£v. el Be auvdexcx;, [i£xpf|a£i tic; auxov dpid^ioc;. [lExpdxw, xal eaxw 6 T. xal sitsl 6 T xov B [iexpsT, 6 8s: B xov A [iexpsT, xal 6 T apa xov A [isxpeT. xal si (lev Ttpfixoc; eaxiv 6 T, yeyovoc; av sir) xo ETiixax'Osv. ei 8s auvdexoc;, [iexpf|aei xi? auxov dpi%6<;. xoiauxrjc; 8r) yivo[ievr)<; eiuaxeijiewc; X/jcp-drjaexai xic; TtpGxoc; dpid^toc;, oc; [iexpf|aei. ei yap ou X/jcp-drjoexai, [lexprpouai xov A dpid^iov omeipoi dpid^toi, Sv exepoc exepou eXdaatov eaxiv oitep eaxiv dSuvaxov ev dpnJ^olc;. Xrjcp'driaexai xic; apa Ttpwxoc; dpi'd^xot;, oc; [lexprpei xov Tipo eauxou, oc; xal xov A ^Expfjaei. A' ■ B 1 r> 1 "Auac; apa auvdevxoc; dpid[x6c; utio Ttpcoxou xivoc; dpid^iou [icxpelxai- orcep e5ei SeTc;ai. X(3'. "Aitac; dpid^toc; fjxoi TtpGxoc; eaxiv fj utio upwxou xivoc; dprd^ioO [xexpelxai. ' A' 1 TCaxco dpi%6<; 6 A- Xeyto, oxi 6 A f]xoi Ttpcoxoc; eaxiv fj utio Tipwxou xivoc; dpid^tou [icxpelxai. El [iev ouv Ttpwxoc; eaxiv 6 A, yeyovoc; av €ir\ xo CTiixax'Oev. ei Be auvdexoc;, [iexpr|aei xu; auxov TtpGxoc; dpid \±6c,. "Arcac; apa dpiiiJjjioc; rjxoi TtpGxoc; eaxiv fj utio upwxou xivoc; dpn5[jio0 [xexpelxai- onep e8ei SeT^ai. Xy'. Api-djjiov Bo'devxwv ouoawvouv eupelv xouc; eXaxiaxouc; xwv xov auxov Xoyov exovxcov auxolc;. "Eaxwaav ol SoiSevxec; orcoaoiouv dpid^iol ol A, B, T' 8eT 8r) eupelv xouc; eXaxiaxouc; xwv xov auxov Xoyov exovxcov xou; A, B, T. Ol A, B, r yap fjxoi upfixoi Ttpoc; dXXfjXouc; eialv fj ou. ei [i£v ouv ol A, B, T Tipoxoi rcp6<; dXXrjXouc; claw, eXdxiaxoi eiai xwv xov auxov Xoyov exovxov auxolc;. is prime then that which was prescribed has happened. And if (B is) composite then some number will measure it. Let it (so) measure (B), and let it be C. And since C measures B, and B measures A, C thus also measures A. And if C is prime then that which was prescribed has happened. And if (C is) composite then some number will measure it. So, in this manner of continued inves- tigation, some prime number will be found which will measure (the number preceding it, which will also mea- sure A) . And if (such a number) cannot be found then an infinite (series of) numbers, each of which is less than the preceding, will measure the number A. The very thing is impossible for numbers. Thus, some prime number will (eventually) be found which will measure the (number) preceding it, which will also measure A. A' 1 B 1 O 1 Thus, every composite number is measured by some prime number. (Which is) the very thing it was required to show. Proposition 32 Every number is either prime or is measured by some prime number. A i 1 Let A be a number. I say that A is either prime or is measured by some prime number. In fact, if A is prime then that which was prescribed has happened. And if (it is) composite then some prime number will measure it [Prop. 7.31]. Thus, every number is either prime or is measured by some prime number. (Which is) the very thing it was required to show. Proposition 33 To find the least of those (numbers) having the same ratio as any given multitude of numbers. Let A, B, and C be any given multitude of numbers. So it is required to find the least of those (numbers) hav- ing the same ratio as A, B, and C. For A, B, and C are either prime to one another, or not. In fact, if A, B, and C are prime to one another then they are the least of those (numbers) having the same ratio as them [Prop. 7.22]. 219 ETOIXEIfiN C- ELEMENTS BOOK 7 ABTAEZHOKAM 1 1 1 I I 1 1 1 1 1 I Et 8e ou, eiXiqcp'do xov A, B, T to ^eyiaxov xoivov ^iixpov 6 A, xai oadxic; 6 A exaaxov xov A, B, T jiexpe'i, xoaauxai ^.ovd8e<; eaxoaav ev exdaxo xov E, Z, H. xai exaaxoc apa xov E, Z, H exaaxov xov A, B, T [icxpeT xaxa xa<; ev xo A jiovd8a<;. oi E, Z, H apa xou<; A, B, T ladxi? ^texpouaiv oi E, Z, H apa xoic; A, B, T ev xo auxo Xoyo eiaiv. Xeyo 8rj, oxi xai eXd)(iaxoi. ei yap \ifi eiaiv oi E, Z, H eXd)(iaxoi xov xov auxov Xoyov e)(6vxov xoi<; A, B, T, eaovxai [xivec] xov E, Z, H eXdaaovec; dpid^ioi ev tu auxo Xoyo ovxec xou; A, B, T. eaxoaav oi 0, K, A- iadxu; apa 6 xov A jiexpel xai exdxepo<; xov K, A exdxepov iwv B, T. oadxu; Se 6 xov A ^expeT, xoaauxai ^iovd8e<; eaxoaav ev xo M- xai exdxepo<; apa xov K, A exdxepov xov B, T ^texpei xaxa xa<; ev xo M ^iovd8a<;. xai eitei 6 xov A ^.expeT xaxa xac; ev iu M ^iovd8a<;, xai 6 M apa xov A jiexpei xaxa xa<; ev xo [iovdSa<;. 8ia xa auxa Sr| 6 M xai exdxepov xov B, r jiexpei xaxa xac; ev exaxepo xov K, A ^iovd8ac 6 M apa xou<; A, B, r ^.expeT. xai eitei 6 xov A jiexpei xaxa xa<; ev xo M jiovdSac;, 6 apa xov M 7ioXXan;Xaaidaa<; xov A 7ieTio(r]xev. 8ia xa auxa 8r) xai 6 E xov A TtoXXaTtXaaidaac; xov A 7ieTio(r]xev. iao<; apa eaxiv 6 ex xov E, A xo ex iwv 0, M. eaxiv apa (i><; 6 E 7tp6<; xov 0, ouxo<; 6 M 7tp6<; xov A. ^isi^tov Se 6 E xou ©• ^eii^ov apa xai 6 M xou A. xai ^xexpei xoug A, B, E oitep eaxiv dBuvaxov tmoxeixai yap 6 A xov A, B, T xo (jieyiaxov xoivov (icxpov. oux apa eaovxai xive<; xov E, Z, H eXdaaove<; dpud^oi ev xo auxo Xoyo ovxec; xou; A, B, T. oi E, Z, H apa eXd)(iaxo[ eiai xov xov auxov Xoyov exovxov xoic A, B, E oiiep e8ei Sel^ai. X8'. Auo dpid^ov Bo'devxov eupeiv, ov eXd)(iaxov jiexpouaiv dprd^iov. 'Eaxwaav oi BoiSevxec; 8uo dpnSjioi oi A, B- Sei Sr) eupeiv, ABCDEFGHKLM 1 1 I 1 1 I 1 1 I 1 I And if not, let the greatest common measure, D, of A, B, and C have be taken [Prop. 7.3]. And as many times as D measures A, B, C, so many units let there be in E, F, G, respectively And thus E, F, G mea- sure A, B, C, respectively, according to the units in D [Prop. 7.15]. Thus, E, F, G measure A, B, C (respec- tively) an equal number of times. Thus, E, F, G are in the same ratio as A, B, C (respectively) [Def. 7.20]. So I say that (they are) also the least (of those numbers hav- ing the same ratio as A, B, C). For if E, F, G are not the least of those (numbers) having the same ratio as A, B, C (respectively), then there will be [some] numbers less than E, F, G which are in the same ratio as A, B, C (respectively). Let them be H, K, L. Thus, H measures A the same number of times that K, L also measure B, C, respectively. And as many times as H measures A, so many units let there be in M. Thus, K, L measure B, C, respectively, according to the units in M. And since H measures A according to the units in M, M thus also measures A according to the units in H [Prop. 7.15]. So, for the same (reasons), M also measures B, C accord- ing to the units in K, L, respectively. Thus, M measures A, B, and C. And since H measures A according to the units in M, H has thus made A (by) multiplying M. So, for the same (reasons), E has also made A (by) multiply- ing D. Thus, the (number created) from (multiplying) E and D is equal to the (number created) from (multi- plying) H and M. Thus, as E (is) to H, so M (is) to D [Prop. 7.19]. And E (is) greater than H. Thus, M (is) also greater than D [Prop. 5.13]. And (M) measures A, B, and C. The very thing is impossible. For D was assumed (to be) the greatest common measure of A, B, and C. Thus, there cannot be any numbers less than E, F, G which are in the same ratio as A, B, C (respec- tively). Thus, E, F, G are the least of (those numbers) having the same ratio as A, B, C (respectively) . (Which is) the very thing it was required to show Proposition 34 To find the least number which two given numbers (both) measure. Let A and B be the two given numbers. So it is re- 220 ETOIXEIfiN C- ELEMENTS BOOK 7 ov eXd)(iaTov ^texpouaiv dpi'd^ov. A i 1 B' 1 ri 1 Ai 1 Ei ' Z ' Oi A, B yap f\ioi rcpcoxoi Ttpoc; dXXfjXouc; eiaiv fj ou. eaxcoaav rcpoxepov oi A, B Tipwxoi rcpoc; dXXrjXouc;, xai 6 A xov B TxoXXanXaaidaac; xov T Ttoieixw xdi 6 B apa xov A TioXXomXaaidaoK; xov T 7i£K0ir]xev. oi A, B apoc xov T \is- xpouaiv. Xeyco 8r], Sxi xai sXd)(iaxov. ei yap \±r\, \±£Tpr\aouoi xiva dpii^ov oi A, B iXdaaova ovxa xou T. ^.expeixwaav xov A. xai oadxu; 6 A xov A ^xexpeT, xoaauxai jiovd8e<; eaxwaav ev x£> E, oadxu; 8e 6 B xov A ^expert, xoaauxai |jovd8ec; eaxioaav ev xfi Z. 6 [lev A apa xov E noXka- TtXaaidaac; xov A TteTrairjxev, 6 8e B xov Z TroXXaTiXaaidaac; xov A TCSTtoi/jxev i'aoc; apa eaxiv 6 ex xcov A, E xai ex xwv B, Z. eaxiv apa cl>g 6 A 7tp6<; xov B, ouxgjc; 6 Z npb<z xov E. oi 8e A, B Ttpfixoi, oi 8e Ttpfixoi xai eXd)(iaxoi, oi Se eXd/iaxoi [icxpouai xouc; xov auxov Xoyov e^ovxac; iadxic; o xe (j.£L^cov xov [iciCova xai 6 eXdaawv xov eXdaaova- 6 B apa xov E jiexpe'i, tb<; £it6[i£voc eno^ievov. xai snel 6 A xouc; B, E KoXXajiXaoidoac; xouc; T, A Ttenoi/jxev, eaxiv apa tbc; 6 B Ttpoc; xov E, ouxgjc; 6 T npbq xov A. [lexpe'i 8e 6 B xov E- piexpeT apa xai 6 T xov A 6 ^ei^cov xov eXdaaova- onep eaxiv d8uvaxov. oux apa oi A, B jiexpouai xiva dpidjiov eXdaaova ovxa xou T. 6 T apa eXd/iaxoc; &v Otto xfiv A, B ^expsixai. A i 1 B i 1 Z' 1 Ei ' r i 1 A' 1 Hi 1 ©i 1 Mr) eaxwaav 8f] oi A, B Ttpfixoi Ttpoc; dXXrjXouc;, xai siXricp'dwaav eXd^iaxoi dpi-djioi x«v xov auxov Xoyov EXovxcov xou; A, B oi Z, E- Xaoc, apa eaxiv 6 ex xwv A, E iw quired to find the least number which they (both) mea- sure. A i 1 B i 1 C' 1 Di 1 E i 1 F i 1 For A and B are either prime to one another, or not. Let them, first of all, be prime to one another. And let A make C (by) multiplying B. Thus, B has also made C (by) multiplying A [Prop. 7.16]. Thus, A and B (both) measure C. So I say that (C) is also the least (num- ber which they both measure) . For if not, A and B will (both) measure some (other) number which is less than C. Let them (both) measure D (which is less than C). And as many times as A measures D, so many units let there be in E. And as many times as B measures D, so many units let there be in F. Thus, A has made D (by) multiplying E, and B has made D (by) multiply- ing F. Thus, the (number created) from (multiplying) A and E is equal to the (number created) from (multi- plying) B and F. Thus, as A (is) to B, so F (is) to E [Prop. 7.19]. And A and B are prime (to one another), and prime (numbers) are the least (of those numbers having the same ratio) [Prop. 7.21], and the least (num- bers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20]. Thus, B measures E, as the following (number measuring) the following. And since A has made C and D (by) multi- plying B and E (respectively), thus as B is to E, so C (is) to D [Prop. 7.17]. And B measures E. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) mea- sure some number which is less than C. Thus, C is the least (number) which is measured by (both) A and B. A i 1 B i 1 F i 1 E i 1 C i 1 D i 1 G i 1 H i 1 So let A and B be not prime to one another. And let the least numbers, F and E, have been taken having the same ratio as A and B (respectively) [Prop. 7.33]. 221 ETOIXEIfiN C- ELEMENTS BOOK 7 ex tc5v B, Z. xal 6 A tov E TroXXaTtXaaidaac; tov T tioieltw xal 6 B apa xov Z TTOXXaTtXaaidaac; tov T tcttoi/jxev oi A, B apa tov T ^ETpouaiv. Xeyio 8r), oxi xal eXd/iaTov. si yap [ir|, (ieTprjoouoi Tiva dpii9|ji6v oi A, B sXdaaova ovTa tou r. [iETpsiTcooav tov A. xal oodxig [Lev 6 A tov A [isxpeX, ToaauTai ^ovd8e<; eaTCoaav ev t« H, oodxig Se 6 B tov A (iSTpsT, ToaauTai jiovdSec; saTwaav ev xfi 8. 6 [lev A apa tov H TraXXaTtXaaidaac; tov A TteTronqxsv, 6 8s B tov 9 7ioXXa7iXaaidoa<; tov A 7T£7TO[r)xev. laoc; apa eotIv 6 ex twv A, H tw ex twv B, 0' sgtiv apa 6 A npoc; tov B, outcoc; 6 9 Ttpoc; tov H. u; 8e 6 A 7tpo<; tov B, outioc; 6 Z npog tov E- xal &>c, apa 6 Z Ttpoc; tov E, outgjc; 6 9 upog tov H. oi Bs; Z, E sXd/ioToi, oi 8e sXd/ioToi jiSTpouai tou<; tov auTov Xoyov e^ovTac; iadxic; o te jiei^mv tov jisi^ova xal 6 sXdaawv tov sXdaaova- 6 E apa tov H [iSTpel. xal end 6 A touc; E, H TraXXaTtXaaidaac; touc; T, A kstioi/jxev, ecmv apa &>c, 6 E Ttpoc; tov H, outgjc; 6 T Ttpoc; tov A. 6 Be E tov H [iSTpel' xal 6 T apa tov A ^ETpsI 6 [isi^cov tov sXdaaova- OTtep saw dSuvaTov. oux apa oi A, B [iSTprpouai Tiva dpi-djiov eXdaaova ovTa tou T. 6 T apa sXd/iaToc; d>v utto tGv A, B [jieTpslTai- OTtep enei 5eic;ai. Thus, the (number created) from (multiplying) A and E is equal to the (number created) from (multiplying) B and F [Prop. 7.19]. And let A make C (by) multiplying E. Thus, B has also made C (by) multiplying F. Thus, A and B (both) measure C. So I say that (C) is also the least (number which they both measure) . For if not, A and B will (both) measure some number which is less than C. Let them (both) measure D (which is less than C) . And as many times as A measures D, so many units let there be in G. And as many times as B measures D, so many units let there be in H. Thus, A has made D (by) multiplying G, and B has made D (by) multiplying H. Thus, the (number created) from (multiplying) A and G is equal to the (number created) from (multiplying) B and H. Thus, as A is to B, so H (is) to G [Prop. 7.19]. And as A (is) to B, so F (is) to E. Thus, also, as F (is) to E, so (is) to G. And F and are the least (num- bers having the same ratio as A and B), and the least (numbers) measure those (numbers) having the same ra- tio an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20]. Thus, E measures G. And since A has made C and D (by) mul- tiplying E and G (respectively), thus as E is to G, so C (is) to D [Prop. 7.17]. And E measures G. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) mea- sure some (number) which is less than C. Thus, C (is) the least (number) which is measured by (both) A and B. (Which is) the very thing it was required to show. Xe'. 'Edv 6uo dpi'd^.oi dpid^iov Tiva jiSTpGaiv, xal 6 eXd/iarac; un auTGv (jieTpou^ievoc; tov auTov jiSTprpsi. A i 1 Bi 1 r z a I 1 1 Ei 1 Auo yap dpi%ol oi A, B dpi%6v Tiva tov FA ^e- TpeiToaav, eXd)(iaTov 8e tov E- Xeyw, oti xal 6 E tov TA (iETpeT. Ei yap ou [Lsjpei 6 E tov PA, 6 E tov AZ ^tETpwv Xsitctco eauTou sXdaaova tov TZ. xal ETtri oi A, B tov E ^tSTpoOaiv, 6 8e E tov AZ \izxpei, xal oi A, B apa tov AZ ^i£Tpr]aouaiv. ^ETpouai 8e xal 6Xov tov TA- xal XoiTtov apa tov rZ [iETprjaouaiv eXdaaova ovTa tou E' oitsp sgtIv dSuvaTov. oux apa ou (iSTpsT 6 E tov FA- (JETpsT apa- onep ebei 8eic;ai. Proposition 35 If two numbers (both) measure some number then the least (number) measured by them will also measure the same (number). A i 1 B ' 1 C F D i 1 1 Ei 1 For let two numbers, A and B, (both) measure some number CD, and (let) E (be the) least (number mea- sured by both A and B) . I say that E also measures CD. For if E does not measure CD then let E leave CF less than itself (in) measuring DF. And since A and B (both) measure E, and E measures DF, A and B will thus also measure DF. And (A and B) also measure the whole of CD. Thus, they will also measure the remainder CF, which is less than E. The very thing is impossible. Thus, E cannot not measure CD. Thus, (E) measures 222 ETOIXEIfiN C- ELEMENTS BOOK 7 W. Tpifiv dpi%Gv 8oi3evx«v supeiv, ov eXd)(iaxov [jie- xpouaiv aptduov. TSaxwaav oi BoiDevxEi; xpeu; dpid^ol ol A, B, E 8ei 8r) eOpsTv, ov eXd)(iaxov ^sxpouaiv dpid^iov. A' 1 B i 1 r i 1 a i ' E' 1 Z' ' EiXr ! ](p , d« yap Otto 860 xwv A, B eXd)(iaxoc; ^lexpou^ievoc; 6 A. 6 8f) r xov A rjxoi ^.sxpei f\ ou jisxpei. [lexpsixw jcpoxepov. [isxpouoi 8e xai 01 A, B xov A' oi A, B, T apa xov A jiexpouaiv. Xeyw S^, oxi xal eXd)(iaxov. el yap [ir], [iexpr]aouaiv [xiva] dpid^iov oi A, B, T erXdaaova ovxa xou A. ^expeixwaav xov E. inel oi A, B, T xov E ^.expouaiv, xal oi A, B apa xov E ^xexpouaiv. xal 6 sXd/iaxot; apa 6ti:6 xfiv A, B piexpou^evoc; [xov E] jjiexprpei. eXd)(iaxoc; 8e Otto xGv A, B |iexpou^ev6<; eaxiv 6 A- 6 A apa xov E |iexpf)aei 6 (lei^tov xov sXdaaova' ousp eaxiv dSuvaxov. oux apa oi A, B, T (lexpiqoouoi xiva dpi%6v sXdaaova ovxa xou A- oi A, B, T apa sXd/iaxov xov A jisxpouaiv. Mr) ^lExpeixco Sf| TidXiv 6 T xov A, xai eiXiqcp'dio bub xwv T, A sXd/iaxoc; ^expou^evoc; dpi%6c; 6 E. end oi A, B xov A ^expouaiv, 6 8e A xov E ^expsi, xai oi A, B apa xov E [isxpouaiv. jiexpsi 8e xai 6 T [xov E- xai] oi A, B, T apa xov E jisxpouaiv. Xeyw 8rj, oxi xal sXd/iaxov. si yap [if], |iexpf|aoua[ xiva oi A, B, T eXdooova ovxa xou E. (lexpsixwaav xov Z. sicd oi A, B, T xov Z ^xexpouaiv, xai oi A, B apa xov Z [lexpouaiv xai 6 eXdxioxog apa bub xwv A, B [isxpoujisvog xov Z jiexprjoei. £Xd)(iaxo<; 8e bub xwv A, B ^.expou^evoc; eaxiv 6 A' 6 A apa xov Z jiexpe'i. ^.expeT 8e xal 6 T xov Z - oi A, T apa xov Z ^.expouaiv fiaie xai 6 eXd)(iaxoc; Otto xwv A, T ^iexpou^ievo<; xov Z jiexpr|aei. 6 8e eXd)(iaxo<; (mo xwv T, A ^expou^ievo<; eaxiv 6 E- 6 E apa xov Z ^.expeT 6 [icii^wv xov eXdaaova- ouep eoxiv dSuvaxov. oux apa oi A, B, T ^expr|aoua[ xiva dpid^tov eXdaaova ovxa xou E. 6 E apa eXa^iajoc, wv utio x£Sv A, B, T ^lexpelxai- oTiep e8ei SeT^ai. (CD). (Which is) the very thing it was required to show. Proposition 36 To find the least number which three given numbers (all) measure. Let A, B, and C be the three given numbers. So it is required to find the least number which they (all) mea- sure. A 1 1 B ' ' C ' ' D 1 1 E ' 1 F ' 1 For let the least (number), D, measured by the two (numbers) A and B have been taken [Prop. 7.34]. So C either measures, or does not measure, D. Let it, first of all, measure (D). And A and B also measure D. Thus, A, B, and C (all) measure D. So I say that (D is) also the least (number measured by A, B, and C) . For if not, A, B, and C will (all) measure [some] number which is less than D. Let them measure E (which is less than D). Since A, B, and C (all) measure E then A and B thus also measure E. Thus, the least (number) measured by A and B will also measure [E] [Prop. 7.35]. And D is the least (number) measured by A and B. Thus, D will measure E, the greater (measuring) the lesser. The very thing is impossible. Thus, A, B, and C cannot (all) measure some number which is less than D. Thus, A, B, and C (all) measure the least (number) D. So, again, let C not measure D. And let the least number, E, measured by C and D have been taken [Prop. 7.34]. Since A and B measure D, and D measures E, A and B thus also measure E. And C also measures [E] . Thus, A, B, and C [also] measure E. So I say that (E is) also the least (number measured by A, B, and C) . For if not, A, B, and C will (all) measure some (number) which is less than E. Let them measure F (which is less than E). Since A, B, and C (all) measure F, A and B thus also measure F. Thus, the least (number) measured by A and B will also measure F [Prop. 7.35]. And D is the least (number) measured by A and B. Thus, D measures F. And C also measures F. Thus, D and C (both) measure F. Hence, the least (number) measured by D and C will also measure F [Prop. 7.35]. And E 223 ETOIXEIfiN C- ELEMENTS BOOK 7 'Edv dpi/duo? (mo xivog dpidjiou [lexprjxai, 6 [isxpoujisvog 6(itovu^ov ^£po<; e^si xw fisxpoOvxi. A' 1 B 1 r> 1 A' — ' Apid^xot; yap 6 A tmo xivo<; apiduoO xoO B ^STpeCcrdco - Xsy", oxi 6 A o^iwvu^iov ^xepo<; sxsi xw B. 'Oadxu; yap 6 B xov A [lexpeT, xoaauxai ^.ovd8e;<; eoxw- aav ev xw I\ STtri 6 B xov A [Lexpei xaxd xdc; ev xfi T [LovabaQ, ^lexpel Be xal f) A ^tovdc; xov T dpid^iov xaxd xdc; ev auxC) ^ovdSag, ladxic; dpa f\ A jiovdc; xov T dpi'djiov ^.e- xpeT xal 6 B xov A. evaXXdc; dpa ladxic; f) A \io\>aq xov B dpid^tov ^texpcT xal 6 T xov A- o dpa ^tepoc; eaxlv f) A jiovdc; xoO B dpid^oO, xo auxo ^icpoc; eaxl xal 6 T xou A. r) 8e A jiovdc; xou B dpid^ou \±epoq eaxlv 6^wvu[iov aux£y xal 6 T dpa xou A ^tepoc; eaxlv o^xovu^ov xO B. waxe 6 A ^tepoc; exei xov T o^ovu^tov ovxa iw B- oitep e8ei 8eTc;ai. XT]'. 'Eav dpi-duoc; ^tepoc; e^Tl oxiouv, Otto o^covu^iou dpi%oO piexpr^riaexai xw uepei. 1 B 1 V 1 A> — i Apid^oc; yap 6 A ^.epoc; exexco oxiouv xov B, xal tw B [jiepei o^xwvu^oc; eoxw [dpidjioc;] 6 T- Xeyoj, oxi 6 T xov A ^expel. Tkel yap 6 B xou A \±epoq eaxlv o^wvujiov x£> T, eaxi 8e xal f] A jiovdc; xou T [izpoz o^xovu^ov auxo, o dpa ^iepo<; is the least (number) measured by C and D. Thus, E measures F, the greater (measuring) the lesser. The very thing is impossible. Thus, A, B, and C cannot measure some number which is less than E. Thus, E (is) the least (number) which is measured by A, B, and C. (Which is) the very thing it was required to show. Proposition 37 If a number is measured by some number then the (number) measured will have a part called the same as the measuring (number). A' 1 B' 1 O 1 D' 1 For let the number A be measured by some number B. I say that A has a part called the same as B. For as many times as B measures A, so many units let there be in C. Since B measures A according to the units in C, and the unit D also measures C according to the units in it, the unit D thus measures the number C as many times as B (measures) A. Thus, alternately, the unit D measures the number B as many times as C (measures) A [Prop. 7.15]. Thus, which(ever) part the unit D is of the number B, C is also the same part of A. And the unit D is a part of the number B called the same as it (i.e., a £>th part). Thus, C is also a part of A called the same as B (i.e., C is the £?th part of A). Hence, A has a part C which is called the same as B (i.e., A has a Bth part) . (Which is) the very thing it was required to show. Proposition 38 If a number has any part whatever then it will be mea- sured by a number called the same as the part. A i 1 Bi 1 Ci 1 Di 1 For let the number A have any part whatever, B. And let the [number] C be called the same as the part B (i.e., B is the Cth part of A) . I say that C measures A. For since B is a part of A called the same as C, and the unit D is also a part of C called the same as it (i.e., 224 ETOIXEIfiN C- ELEMENTS BOOK 7 saxlv f] A ^lovac xoO T dpid^iou, to auxo [iepo<; Sail xal 6 B xoO A- tadxu; apa rj A ^.ova<; xov T dpid^iov ^.expeT xal 6 B xov A. evaXXd^ apa iadxu; f] A ^tovac; xov B dpid^iov ^expei xal 6 T xov A. 6 T apa xov A ^xexpsi- oTtep eSei Bd^ai. Aft'. Apuduov eupefv, o<; eXd)(iaxo<; wv e^ei xa Scydevxa ^tcpr]. a b r i — i i 1 i 1 D is the Cth part of C), thus which (ever) part the unit D is of the number C, B is also the same part of A. Thus, the unit D measures the number C as many times as B (measures) A. Thus, alternately, the unit D measures the number B as many times as C (measures) A [Prop. 7.15] . Thus, C measures A. (Which is) the very thing it was required to show. Proposition 39 To find the least number that will have given parts. ABC i — i i 1 i 1 A E i 1 i ■ D ■ , H H Tiaxco xa So'dsvxa y.epf] xa A, B, T- 8d Br] apid^ov supeiv, 6<z eXd)(iaxoc; <J>v s^ei xa A, B, T jiepr). "Eaxwaav yap xou; A, B, T ^ispeaiv o^covu^ioi dpid^tol oi A, E, Z, xal dXrjcp'dw (mo xGv A, E, Z 6Xd)(iaxo<; \±s- xpoujievoc; dprdjioi; 6 H. c O H apa o^tcovupia [ispf] £X £l xo ^ A, E, Z. xou; Be A, E, Z o^covu^ia [ispr) eaxl xa A, B, T- 6 H apa ex £l T & A, B, T (ieprj. Xsyw 8rj, oxi xal eXd)(iaxo<; wv, eE yap ^V), eaxai xu; xoO H eXdaowv dpnf)uo<;, o<; e^a xa A, B, T [ispr\. eaxw 6 6. etieI 6 6 ex £l ™ A, B, T fiepr], 6 6 apa bub o^covu^cov dpi-d^cov ^t£xpr]Tf)r]0£xai xou; A, B, T [ispeaiv. xou; Be A, B, r \iepeow b\i(x>wy.oi dpn^ot daiv oi A, E, Z- 6 9 apa utco xwv A, E, Z jiexpdxai. xa[ eaxiv eXdaawv xou H- orcsp eaxlv dBuvaxov. oux apa eaxai xu; xou H iXdaatov dpudu6<;, be, e^si xa A, B, T [iepf]- bnep eBei BsT^ai. Let A, B, and C be the given parts. So it is required to find the least number which will have the parts A, B, and C (i.e., an Ath part, a Bth part, and a Cth part). For let D, E, and F be numbers having the same names as the parts A, B, and C (respectively). And let the least number, G, measured by D, E, and F, have been taken [Prop. 7.36]. Thus, G has parts called the same as D, E, and F [Prop. 7.37]. And A, B, and C are parts called the same as D, E, and F (respectively). Thus, G has the parts A, B, and C. So I say that (G) is also the least (number having the parts A, B, and C). For if not, there will be some number less than G which will have the parts A, B, and C. Let it be H . Since H has the parts A, B, and C, H will thus be measured by numbers called the same as the parts A, B, and C [Prop. 7.38]. And D, E, and F are numbers called the same as the parts A, B, and C (respectively). Thus, H is measured by D, E, and F. And (H) is less than G. The very thing is impossible. Thus, there cannot be some number less than G which will have the parts A, B, and C. (Which is) the very thing it was required to show. 225 226 ELEMENTS BOOK 8 Continued Proportion tThe propositions contained in Books 7-9 are generally attributed to the school of Pythagoras. 227 STOIXEIQN rf. ELEMENTS BOOK 8 a . 'Eav Saiv oaoiSrjuoxouv apid^ol l^fjc; dvdXoyov, oi 8e dxpoi auxfiv Ttpfixoi Tip6<; dXXrjXout; waiv, eXd)(iaxoi eiai x£Sv xov auxov Xoyov sxovxcov auxolc ' Ei — ' B ' ' Z ' T ' ' H' ' A i 1 ®i 1 TCaxcoaav oTtoaoiouv dpi-d^ol e^r|<; dvdXoyov oi A, B, r, A, oi Be dxpoi auxaiv oi A, A, Ttpaixoi Ttpoc; aXkr\kovc, eaxcoaav Xeyio, oxi oi A, B, T, A eXd)(iaxoi eiai xfiv tov auxov Xoyov e/ovxiov auxolc;. Ei yap [jltj, eaxwaav eXdxxovec; xaiv A, B, T, A oi E, Z, H, ev xw auxcS Xoyio ovxec; auxolc;. xal end oi A, B, T, A ev xtp auxcp Xoycp etal xolc; E, Z, H, 0, xod eaxiv i'oov xo KkrjdoQ [x«v A, B, T, A] x£> TiXri'dei [x«v E, Z, H, 9], 8i' i'oou dpa eaxlv <i>c; 6 A upog xov A, 6 E Ttpoc; xov 9. oi Be A, A ixpwxoi, oi Bs itpcoxoi xal eXd)(iaxoi, oi 6e eXd/iaxoi dpidjiol ^.expouai xouc; xov auxov Xoyov e^ovxac; iadxic; o xe (iei^cov xov (jiei^ova xai 6 eXdaawv xov eXdaaova, xouxeaxiv o xe fjyou^ievoc xov fjyou^ievov xal 6 en6[ievoc, xov E7i6|jievov. jiexpel dpa 6 A xov E 6 [iei^cov xov eXdaaova' oTtep eaxiv dSuwxov. oux dpa oi E, Z, H, 9 eXdaaovec; ovxec; xwv A, B, T, A sv x£5 auxw X6ya> eiaiv auxolc;. oi A, B, r, A dpa eXd)(iaxoi eiai xfiv xov auxov Xoyov e/ovxiov auxolc;- oTtep eSei Belial. P'- Aptduouc; eupeiv ec;rjc; dvdXoyov eXa)([axouc;, oaouc; dv eraxd^r) xic;, ev iu SoiSevxi Xoyw. TEaxco 6 Bo'delc; Xoyoc; ev eXd)(iaxoic; dpn&iio'ic; 6 xou A rcpoc; xov B' Bel 8r) dpid^ouc; eupeiv ec;rjc; dvdXoyov eXa/iaxouc;, oaouc; dv xic imia£,r\, sv x« xou A Ttpoc; xov B Xoycp. 'Eiuxexdx'dwaav 8rj xeaaapec;, xal 6 A eauxov uoXXa- TtXaaidaac; xov T Tioieixw, xov Be B TioXXauXaaidaac; xov A Ttoieixw, xal exi 6 B eauxov noXXaTiXaaidaac; xov E ttoieixgj, xal exi 6 A xouc; T, A, E TtoXXaitXaaidaac; xouc; Z, H, 9 Tioieixw, 6 Be B xov E TtoXXaTtXaaidaac; xov K Ttoieixw. Proposition 1 If there are any multitude whatsoever of continuously proportional numbers, and the outermost of them are prime to one another, then the (numbers) are the least of those (numbers) having the same ratio as them. A i ' E i 1 B i 1 F i 1 C i 1 G' 1 D ' H' ' Let A, B, C, D be any multitude whatsoever of con- tinuously proportional numbers. And let the outermost of them, A and D, be prime to one another. I say that A, B, C, D are the least of those (numbers) having the same ratio as them. For if not, let E, F, G, H be less than A, B, C, D (respectively), being in the same ratio as them. And since A, B, C, D are in the same ratio as E, F, G, H, and the multitude [of A, B, C, D] is equal to the multitude [of E, F, G, H], thus, via equality as A is to D, (so) E (is) to H [Prop. 7.14]. And A and D (are) prime (to one another). And prime (numbers are) also the least of those (numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser — that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, A measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, E, F, G, H, being less than A, B, C, D, are not in the same ratio as them. Thus, A, B, C, D are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show. Proposition 2 To find the least numbers, as many as may be pre- scribed, (which are) continuously proportional in a given ratio. Let the given ratio, (expressed) in the least numbers, be that of A to B. So it is required to find the least num- bers, as many as may be prescribed, (which are) in the ratio of A to B. Let four (numbers) have been prescribed. And let A make C (by) multiplying itself, and let it make D (by) multiplying B. And, further, let B make E (by) multiply- ing itself. And, further, let A make F, G, H (by) mul- tiplying C, D, E. And let B make K (by) multiplying E. 228 STOIXEIQN rf. ELEMENTS BOOK 8 a i 1 r i 1 B i 1 A i 1 Ei 1 Z' ' Hi ' ©i 1 K 1 Kal tnel 6 A eauxov [lev TtoXXaTtXaaidaae xov T 7i£iio(r]X£v, tov 8e B TtoXXaTtXaaidaae xov A kstioitjxev, eaxiv apa u<;oA Ttpoe xov B, [ouxcoe] 6 T npbc, xov A. TidXiv, ercel 6 [lev A xov B TtoXXauXaaidaae xov A TteTtoi/jxev, 6 Se B eauxov TtoXXaitXaaidaae xov E 7i£no(r]xev, exdxepoe apa tuv A, B xov B rcoXXanXaaidaae exdxepov x(5v A, E 7i£iio(r]xev. eaxiv apa (be 6 A Ttpoe xov B, ouxwe 6 A Ttpoe xov E. dXX'' (be 6 A Ttpoe xov B, 6 T Ttpoe xov A- xal (be apa 6 T Ttpoe xov A, 6 A Ttpoe xov E. xal CTtel 6 A xoue T, A TtoXXaTtXaaidaae xoue Z, H TteTtoirjxev, eaxiv apa (be 6 T Ttpoe xov A, [ouxw<;] 6 Z Ttpoe xov H. (be 8e 6 T Ttpoe xov A, ouxwe rjv 6 A Ttpoe xov B- xal (be apa 6 A Ttpoe xov B, 6 Z Ttpoe xov H. TtdXiv, CTtel 6 A xouc A, E TtoXXaTtXaaidaae xouc H, TteTtoirjxev, eaxiv apa (be 6 A Ttpoe xov E, 6 H Ttpoe xov 9. dXX' (be 6 A Ttpoe xov E, 6 A Ttpoe xov B. xal (be apa 6 A Ttpoe xov B, ouxwe 6 H Ttpoe xov 0. xal excel oi A, B xov E TtoXXaitXaaidaavxee xouc 9, K TteTtoifjxaaiv, eaxiv apa (be 6 A Ttpoe xov B, ouxwe 6 9 Ttpoe xov K. dXX'' (be 6 A Ttpoe xov B, ouxwe 6 xe Z Ttpoe xov H xal 6 H Ttpoe xov 9. xal (be apa 6 Z Ttpoe xov H, ouxwe o xe H Ttpoe xov 9 xal 6 9 Ttpoe xov K- oi T, A, E apa xal oi Z, H, 9, K dvdXoyov eiaiv ev x£> xou A Ttpoe xov B Xoyw. Xeyw 8r), 6xi xal eXd)(iaxoi. eitel yap oi A, B eXd)(iaxo( eiai xwv xov auxov Xoyov e)(6vxwv auxole, oi Be eXd)(iaxoi xwv xov auxov Xoyov £)(6vxwv Ttpwxoi Ttpoe dXXf|Xoue eiaiv, oi A, B apa upwxoi Ttpoe dXXrjXouc eiaiv. xal exdxepoe \±kv xwv A, B eauxov TtoXXaTtXaaidaae exdxepov xwv T, E TtCTtoir)xev, exdxepov 8e x£>v T, E TtoXXaTtXaaidaae exdxepov xwv Z, K Tteito(r]xev oi T, E apa xal oi Z, K TtpGxoi Ttpoe dXXr|Xoue eiaiv. eav Se waiv oTtoaoioOv dpid^oi e£/je dvdXoyov, oi 8e axpoi auxCSv Ttpwxoi Ttpoe dXXr]Xoue Saiv, eXd)(iaxo[ eiai xwv xov auxov Xoyov exovxov auxole- oi T, A, E apa xal oi Z, H, 9, K eXdxiaxoi eiai xwv xov auxov Xoyov exovxeov xou; A, B- oTtep eSei Bel^ai. A i 1 O 1 B i 1 D 1 Ei 1 F i 1 Gi 1 Hi 1 Ki 1 And since A has made G (by) multiplying itself, and has made D (by) multiplying B, thus as A is to B, [so] G (is) to D [Prop. 7.17]. Again, since A has made D (by) multiplying B, and B has made E (by) multiplying itself, A, B have thus made D, E, respectively, (by) multiplying B, Thus, as A is to B, so D (is) to E [Prop. 7.18]. But, as A (is) to B, (so) C (is) to D. And thus as G (is) to D, (so) D (is) to E. And since A has made G (by) multiplying C, D, thus as C is to L>, [so] F (is) to G [Prop. 7.17]. And as G (is) to D, so A was to B. And thus as A (is) to B, (so) F (is) to G. Again, since A has made G, (by) multiplying I?, E, thus as £> is to E, (so) G (is) to H [Prop. 7.17]. But, as D (is) to £, (so) A (is) to B. And thus as A (is) to B, so G (is) to H. And since A, B have made AT (by) multiplying E, thus as A is to B, so H (is) to AT. But, as A (is) to B, so F (is) to G, and G to B. And thus as F (is) to G, so G (is) to B, and H to A". Thus, G, D, E and B, G, B, AT are (both continu- ously) proportional in the ratio of A to B. So I say that (they are) also the least (sets of numbers continuously proportional in that ratio). For since A and B are the least of those (numbers) having the same ratio as them, and the least of those (numbers) having the same ratio are prime to one another [Prop. 7.22], A and B are thus prime to one another. And A, B have made G, B, respec- tively, (by) multiplying themselves, and have made F, K by multiplying G, B, respectively. Thus, G, E and F, K are prime to one another [Prop. 7.27]. And if there are any multitude whatsoever of continuously proportional numbers, and the outermost of them are prime to one another, then the (numbers) are the least of those (num- bers) having the same ratio as them [Prop. 8.1]. Thus, G, D, E and F, G, H, K are the least of those (continuously proportional sets of numbers) having the same ratio as A and B. (Which is) the very thing it was required to show. 229 STOIXEIQN rf. ELEMENTS BOOK 8 H6pio\j.a. 'Ex Br) xouxou cpavepov, oxi edv xpelc; dpi/d^ioi zZ,ffc dvdXoyov eXd/iaxoi Goi xfiv xov auxov Xoyov e/ovxiov auxolg, oi axpov auxfiv xexpdytovoi eiaiv, edv 8e xeaaapec, xu[3oi. y'- 'Edv fiaiv otioooiouv apid^ol ec;r]<; dvdXoyov eXd)(iax- 01 xfiv xov auxov Xoyov C)(6vxmv auxolc;, oi dxpoi auxtov Tipwxoi upoc dXXrjXouc; eiaiv. A B r A A M N IO 'Eaxtoaav oitoaoiouv dpid^toi E^fj? dvdXoyov eXd/iaxoi xwv xov auxov Xoyov e)(6vx«v auxolg oi A, B, T, A- Xeyio, 6xi oi dxpoi auxwv oi A, A TtpGxoi npbc, dXXrjXouc; eiaiv. EiXf](p , dwaav yap 8uo ^iev apidjiol eXd)(iaxoi ev xw xCSv A, B, T, A Xoyw oi E, Z, xpsT? 6s oi H, 0, K, xod s^fjc; svi TtXeiout;, ecoc; xo Xa|ipav6|i£vov TtXrydoc; 'iaov yevrjxai xw TtXyydei xwv A, B, T, A. eiXf|Cp , d«aav xal eaxwaav oi A, M, N,H. Kai etiei oi E, Z sXd)(iaxoi eiai xwv xov auxov Xoyov e)(ovxa>v auxou;, itpcoxoi npbz dXXrjXouc; eiaiv. xal Excel exdxepov iSv E, Z eauxov ^.ev TioXXaitXaaidaac; exdxepov xuv H, K TtC7ioir)xev, exdxepov 8e x£Sv H, K uoXXa- TiXaaidaac; exdxepov xwv A, H KCTioirjxev, xai oi H, K dpa xai oi A, H npCSxoi npbq dXXr|Xou<; eiaiv. xai enel oi A, B, T, A eXd)(iaxoi eiai xwv xov auxov Xoyov exovxwv auxou;, eiai Be xai oi A, M, N, S eXd^iaxoi ev xw auxcS Xoyw ovxec xdtc A, B, T, A, xai eaxiv laov xo TtXrydoc; xwv A, B, T, A x« TiXr) , dei xwv A, M, N, S, exaaxo<; dpa xwv A, B, T, A exdaxw xCSv A, M, N, 5 i'ao<; eaxiv i'ao<; dpa eaxiv 6 \±ev A x« A, 6 8e A xw S. xai eiaiv oi A, H upwxoi 7ipo<; dXXf]Xou<;. xai oi A, A dpa TtpGxoi npoc, dXXrjXouc; eiaiv oTiep eBei SeTc;ai. Corollary So it is clear, from this, that if three continuously pro- portional numbers are the least of those (numbers) hav- ing the same ratio as them then the outermost of them are square, and, if four (numbers), cube. Proposition 3 If there are any multitude whatsoever of continu- ously proportional numbers (which are) the least of those (numbers) having the same ratio as them then the outer- most of them are prime to one another. A' 1 Ei 1 G' 1 B 1 F ' —i D L M N O Let A, B, C, D be any multitude whatsoever of con- tinuously proportional numbers (which are) the least of those (numbers) having the same ratio as them. I say that the outermost of them, A and D, are prime to one another. For let the two least (numbers) E, F (which are) in the same ratio as A, B, C, D have been taken [Prop. 7.33]. And the three (least numbers) G, H, K [Prop. 8.2]. And (so on), successively increasing by one, until the multitude of (numbers) taken is made equal to the multitude of A, B, C, D. Let them have been taken, and let them be L, M, N, O. And since E and F are the least of those (numbers) having the same ratio as them they are prime to one an- other [Prop. 7.22] . And since E, F have made G, K, re- spectively, (by) multiplying themselves [Prop. 8.2 corn], and have made L, O (by) multiplying G, K, respec- tively, G, K and L, O are thus also prime to one another [Prop. 7.27]. And since A, B, C, D are the least of those (numbers) having the same ratio as them, and L, M, N, O are also the least (of those numbers having the same ratio as them), being in the same ratio as A, B, C, D, and the multitude of A, B, C, D is equal to the multitude of L, M, N, O, thus A, B, C, D are equal to L, M, N, O, respectively. Thus, A is equal to L, and D to O. And L and O are prime to one another. Thus, A and D are also prime to one another. (Which is) the very thing it was 230 STOIXEIQN rf. ELEMENTS BOOK 8 5'. Aoywv SoiSevxov oTioatovouv ev eXayJaxoic; dpid^iou; dpi%ou<; eupelv e^fjc; dvdXoyov eXayJaxouc; ev xoT<; Bo'deTai Xoyou;. ' B r i 1 a E' ' Z N 1 ' © Hi ' H M' ' K O' ' A "Eaxwaav oi SoiSevxec Xoyoi ev eXa)(iaxoic; dpi-d^iou; 6 xe xoO A Tipoc; xov B xal 6 xou T Tip6<; xov A xod exi 6 xoO E npoc, xov Z- 8eT 5r) dpid^iouc; eupelv e^rjc; dvdXoyov eXayJaxouc; ev xe iS xou A Tipoc; xov B Xoytp xal ev xw xou r npoc xov A xal exi iu xou E Tipoc; xov Z. EiXVjcp'dcd yap 6 utio xov B, T eXdyiaxoc; [iexpou^evo<; dpi%6<; 6 H. xal oadxi? y.ev 6 B xov H ^xexpeT, xoaauxdxic xal 6 A xov 6 ^texpeixw, oadxn; 8e 6 T xov H ^texpel, xo- aauxdxic xal 6 A xov K jiexpeixw. 6 Se E xov K fjxoi ^texpel f\ ou jiexpel. ^texpelxw Tipoxepov. xal oadxu; 6 E xov K fie- xpeT, xoaauxdxic xal 6 Z xov A piexpeixo. xal CTtei iadxic; 6 A xov 9 ^texpeT xal 6 B xov H, eaxiv apa ox 6 A Tipoc; xov B, ouxw<; 6 9 Tipoc xov H. Sid xa auxa Srj xal ob<; 6 T Tipoc; xov A, ouxgk 6 H Ttpoc xov K, xal exi foe, 6 E Tipoc; xov Z, ouxoc; 6 K Tipoc xov A- oi 9, H, K, A apa e^fjc dvdXoyov eiaiv ev xe xw xou A Tipoc xov B xal ev iu xou T Tipoc xov A xal exi ev iS xou E Tipoc xov Z Xoyw. Xeyw 8Vj, oxi xal eXdyiaxoi. ei yap eiaiv oi 9, H, K, A ec;rjc dvdXoyov eXdyiaxoi ev xe xolc xou A Tipoc xov B xal xou T Tipoc xov A xal ev x£> xou E Tipoc xov Z Xoyoic, eaxoaav oi N, S, M, O. xal ckci eaxiv 6?6A Tipoc xov B, ouxwc 6 N Tipoc xov S, oi 8e A, B eXd/iaxoi, oi 8e eXdyiaxoi ^expouai xouc xov auxov Xoyov eyovxac iadxic o xe ^ie[£(x>v xov ^.ei^ova xal 6 eXdaaov xov eXdaaova, xouxeaxiv o xe fjyoujievoc xov f)You[ievov xal 6 CTio^evoc xov euo^evov, 6 B apa xov 5 ^.expel. 8ia xa auxa 8r) xal 6 T xov 5 fjiexpeT - oi B, T apa xov S ^.expouaiv xal 6 eXdyiaxoc apa utio twv B, T ^expou^evoc xov H [isipr\asi. eXdyiaxoc Se utio xQv B, T ^.expeTxai 6H'6H apa xov H ^xexpeT 6 ^ei^wv xov eXdaaova - OTiep eaxiv d8uvxaxov. oux apa eaovxai xivec xov 9, H, K, A eXdaaovec dpid^oi ec;rjc ev xe xw xou A Tipoc xov B xal tw xou T Tipoc xov A xal exi tu xou E Tipoc xov Z XoyG. required to show. Proposition 4 For any multitude whatsoever of given ratios, (ex- pressed) in the least numbers, to find the least numbers continuously proportional in these given ratios. A i ' B ' ' C' 1 D 1 E ' ' F ' 1 Ni ' H' ' O' ' G Mi 1 Ki 1 P i 1 L i 1 Let the given ratios, (expressed) in the least numbers, be the (ratios) of A to B, and of C to D, and, further, of E to F. So it is required to find the least numbers continuously proportional in the ratio of A to B, and of C to B, and, further, of E to F. For let the least number, G, measured by (both) B and C have be taken [Prop. 7.34]. And as many times as B measures G, so many times let A also measure H . And as many times as C measures G, so many times let D also measure K. And E either measures, or does not measure, K. Let it, first of all, measure (K). And as many times as E measures K, so many times let F also measure L. And since A measures H the same number of times that B also (measures) G, thus as A is to B, so H (is) to G [Def. 7.20, Prop. 7.13]. And so, for the same (reasons), as C (is) to D, so G (is) to K, and, further, as E (is) to F, so K (is) to L. Thus, H, G, K, L are continuously proportional in the ratio of A to B, and of C to D, and, further, of E to F. So I say that (they are) also the least (numbers con- tinuously proportional in these ratios) . For if H , G, K, L are not the least numbers continuously proportional in the ratios of A to B, and of C to D, and of E to F, let N, O, M, P be (the least such numbers) . And since as A is to B, so N (is) to O, and A and B are the least (numbers which have the same ratio as them), and the least (num- bers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measur- ing) the greater, and the lesser the lesser — that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20], B thus measures O. So, for the same (reasons), C also measures O. Thus, B and C (both) measure O. Thus, the least number measured by (both) B and C will also measure O [Prop. 7.35]. And G (is) the least number measured by (both) B and C. 231 STOIXEIQN rf. ELEMENTS BOOK 8 A' ' B ' T ' ' A ' E' ' Z ' N' ' ®| ' Hi ' H' ' Mi - ' K - ' () ' n> 1 p i 1 s | 1 T 1 Mrj ^sxpsixw 8r) 6 E xov K, xal siXrjcp'dw utio xcov E, K sA&xicttoc; \±£Tpou\±svoz dpid^ioc; 6 M. xal oadxic; \iev 6 K xov M ^sxpsT, xoaauxdxic; xal sxdxspoc; xwv 0, H sxdxspov xwv N, S ^.sxpeixw, oadaxic; 8s 6 E xov M \±e- xpeT, xoaauxdxic; xal 6 Z xov O ^.expeixw. etc! iadxic; 6 9 xov N ^.sxpsT xal 6 H xov S, saxiv apa 6? 6 9 Tipoc; xov H, ouxwc; 6 N Tipoc; xov S. 6c; 8s 6 Tipoc; xov H, ouxcoc 6 A Tipoc; xov B - xal £><; apa 6 A Tipoc; xov B, ouxwc; 6 N Tipoc; xov S. 8id xa auxa hf] xal 6c; 6 T Tipoc; xov A, ouxoc; 6 S Tipoc; xov M. udXiv, etce! iadxic; 6 E xov M ^sxpsl xal 6 Z xov O, saxiv apa wcoE Tipoc; xov Z, ouxoc; 6 M Tipoc; xov 0- oi N, S, M, O apa e^r\c, dvdXoyov siaiv sv xolc; xou xe A Tipoc; xov B xal xou T Tipoc; xov A xal sxi xou E Ttpoc xov Z Xoyou;. Xsyto 8rj, oxi xal sXd)(iaxoi ev xou; A B, T A, E Z Xoyou;. ei yap y.r\, saovxai xivsc; xGv N, S, M, sXdaaovsc; dpi'dpioi Ec^fjc; dvdXoyov ev xou; A B, T A, E Z Xoyou;. Eaxwaav oi n, P, S, T. xal stisi saxiv «<; 6 H Tipoc; xov P, ouxcx; 6 A Tipoc; xov B, oi 8s A, B sXd)(iaxoi, oi 8s sXd)(iaxoi jisxpouai xouc; xov auxov Xoyov S)(ovxac; auxolc; iadxic; o xe r]you^.svo<; xov f]you^isvov xal 6 stiojisvoc; xov £Ti6[i£vov, 6 B apa xov P ^sxpsT. 8ia xa auxa 8r] xal 6 T xov P ^ExpsT- oi B, T apa xov P ^tsxpouaiv. xal 6 sXdxiaxoc; apa utio x<3v B, T jisxou^svoc; xov P ^sxpr|asi. sXdxiaxoc; 8s utio x<3v B, r (jisxpoujjiEvoc; saxiv 6 H- 6 H apa xov P ^ExpsT. xai saxiv cbc; 6 H Tipoc; xov P, ouxwc; 6 K Tipoc; xov £• xal 6 K apa xov S ^sxpsl. [isxpsl 8s xal 6 E xov £• oi E, K apa xov E [isxpouaiv. xal 6 sXdxiaxoc; apa utio xwv E, K ^sxpou^tsvoc; xov S jiExpr]OEi. sXdxiaxoc; 8s utio xGv E, K [isxpou^svoc; saxiv 6 M- 6 M apa xov E [isxpsT 6 ^isiCwv xov sXdaaova- oTisp saxiv d8uvaxov. oux apa saovxai xivsc; xwv Thus, G measures O, the greater (measuring) the lesser. The very thing is impossible. Thus, there cannot be any numbers less than H, G, K, L (which are) continuously (proportional) in the ratio of A to B, and of C to D, and, further, of E to F. A' 1 B Ci 1 D Ei ' F Ni ' H Oi G Mi ' K pi 1 Q R S ' ' T ' So let E not measure K. And let the least num- ber, M, measured by (both) E and K have been taken [Prop. 7.34] . And as many times as K measures M, so many times let H, G also measure N, O, respectively. And as many times as E measures M, so many times let F also measure P. Since H measures N the same num- ber of times as G (measures) O, thus as H is to G, so N (is) to O [Dei. 7.20, Prop. 7.13]. And as H (is) to G, so A (is) to B. And thus as A (is) to B, so N (is) to O. And so, for the same (reasons), as C (is) to D, so O (is) to M. Again, since E measures M the same num- ber of times as F (measures) P, thus as E is to F, so M (is) to P [Def. 7.20, Prop. 7.13]. Thus, N, O, M, P are continuously proportional in the ratios of A to B, and of C to D, and, further, of E to F. So I say that (they are) also the least (numbers) in the ratios of A B, C D, E F. For if not, then there will be some numbers less than N, O, M, P (which are) continuously proportional in the ratios of A B, C D, E F. Let them be Q, R, S, T. And since as Q is to R, so A (is) to B, and A and B (are) the least (numbers having the same ratio as them), and the least (numbers) measure those (numbers) hav- ing the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20], B thus measures R. So, for the same (reasons), C also measures R. Thus, B and C (both) measure R. Thus, the least (number) measured by (both) B and C will also measure R [Prop. 7.35]. And G is the least number measured by (both) B and C. Thus, G measures R. And as G is to R, so K (is) to S. Thus, 232 STOIXEIQN rf. ELEMENTS BOOK 8 N, S, M, O sXdaaovs<; dpi/d^iol s^fjc dvdXoyov sv xs xoTc toO A 7ip6<; xov B xal xoO T Ttpoc; xov A xal exi xoO E 7tp6<; xov Z Xoyoic oi N, S, M, O dpa e^fjc; dvdXoyov eXd)(iaxo( siaiv sv xou; A B, T A, E Z Xoyoic oitsp s8si SsT^ai. £'. Oi STUTts8oi dpiif)[!ot 7tp6<; dXXr|Xou<; Xoyov s/ouai xov auyxsi^isvov sx xwv TtXsupwv. A' 1 B ' ' r i 1 a i 1 E' ' Z ' Hi ' ©| 1 Kj 1 A' ' "Earaaav stutcs8oi apid^ol oi A, B, xal xou [ikv A TtXsupal saxcoaav oi T, A dpi , d[io[, xoO 8s B oi E, Z - Xsya>, oxi 6 A 7ip6<; xov B Xoyov sxsi T ° v cruyxsi^isvov ex xwv TiXeupwv. Aoywv yap BoiDevxwv xoO xe ov eyei 6 T 7tp6<; xov E xal 6 A Ttpoc xov Z eiXfjcp-dwaav dpid^oi s^<; sXd)(iaxoi sv xou; r E, A Z Xoyoic, oi H, 6, K, &axs slvai &><; ^isv xov T npbc, xov E, ouxw<; xov H Ttp6<; xov 0, <i><; Ss xov A 7tp6<; xov Z, ouxcoc; xov Ttpoc; xov K. xal 6 A xov E TtoXXaTcXaaidaac; XOV A TIOISIXCO. Kal lnei 6 A xov jisv T TroXXajcXaaidaac; xov A 7ien;oir]X£v, xov Be E TioXXaTcXaaidaac; xov A iiETco[r)XSv, saxiv dpa 6 T npoc, xov E, oux«<; 6 A upoc; xov A. ci><; 8s 6 r icpoc; xov E, ouxwc; 6 H icpoc; xov ©• xal cb<; dpa 6 H Ttpoc; xov 0, ouxw<; 6 A Ttpoc; xov A. TtdXiv, STtsl 6 E xov A TtoXXaTtXaaidaac; xov A TiETioirjXEv, dXXa ^trjv xal xov Z TtoXXaTtXaaidaac; xov B TtSTtofyxsv, saxiv dpa u; 6 A Ttpoc; xov Z, ouxw<; 6 A Ttpoc; xov B. dXX' cbc; 6 A Ttpoc; xov Z, ouxwc; 6 Ttpoc; xov K- xal cbc; dpa 6 Ttpoc; xov K, ouxox 6 A Ttpoc; xov B. eSsi^i}/] Se xal <i>c; 6 H Ttpoc; xov 0, ouxox 6 A Ttpoc; xov A- 8i° laou dpa saxiv cbc; 6 H Ttpoc; xov K, [ouxck] 6 A Ttpoc; xov B. 6 8s H Ttpoc; xov K Xoyov sxsi if also measures S [Def. 7.20]. And E also measures S 1 [Prop. 7.20]. Thus, E and K (both) measure 5. Thus, the least (number) measured by (both) E and K will also measure S [Prop. 7.35]. And M is the least (number) measured by (both) E and K. Thus, M measures S, the greater (measuring) the lesser. The very thing is impos- sible. Thus there cannot be any numbers less than N, O, M, P (which are) continuously proportional in the ratios of A to B, and of C to D, and, further, of E to F. Thus, N, O, M, P are the least (numbers) continuously propor- tional in the ratios of A B, C D, E F. (Which is) the very thing it was required to show. Proposition 5 Plane numbers have to one another the ratio compoun- ded* out of (the ratios of) their sides. A ' B C 1 Di 1 E i 1 F i 1 G ' Hi ' K 1 L ' ' Let A and B be plane numbers, and let the numbers C, D be the sides of A, and (the numbers) E, F (the sides) of B. I say that A has to B the ratio compounded out of (the ratios of) their sides. For given the ratios which C has to E, and D (has) to F, let the least numbers, G, H, K, continuously propor- tional in the ratios C E, D F have been taken [Prop. 8.4], so that as C is to E, so G (is) to H, and as D (is) to F, so H (is) to K. And let D make L (by) multiplying E. And since D has made A (by) multiplying C, and has made L (by) multiplying E, thus as C is to E, so A (is) to L [Prop. 7.17]. And as C (is) to E, so G (is) to H. And thus as G (is) to H, so A (is) to L. Again, since E has made L (by) multiplying D [Prop. 7.16], but, in fact, has also made B (by) multiplying F, thus as D is to F, so L (is) to B [Prop. 7.17]. But, as D (is) to F, so H (is) to K. And thus as H (is) to K, so L (is) to B. And it was also shown that as G (is) to H, so A (is) to L. Thus, via equality, as G is to K, [so] A (is) to B [Prop. 7.14]. And G has to K the ratio compounded out of (the ratios of) the sides (of A and B). Thus, A also has to B the ratio compounded out of (the ratios of) the sides (of A and B). 233 STOIXEIQN rf. ELEMENTS BOOK 8 xov auyxei^ievov ex xSv TtXeupfiv xod 6 A apex Ttpoc; xov B Xoyov styei xov auyxeijievov ex xfiiv TtXeupfiv ojtep e8ei 8eT<;ai. t i.e., multiplied. f '. 'Edv Saw ottoooiouv dtpi'djiol e<;rj<; dvdXoyov, 6 8e Tipoxoc; xov 8euxepov [ir] nsxpfj, oi)8e aXkoq ouSei? oi)8eva Hexprpei. A' ' B ' ' r ' ' A 1 ' Ei 1 Z' ' H' 1 ©i 1 "Eaxwaav oitoaoioOv dpid^ioi e^rjc; dvdXoyov oi A, B, r, A, E, 6 8e A xov B \xr] jjiexpsixw Xeyw, oxi ouSe aXXoc ou8ek; ouSeva [lexprjaei. "Oxi (iev ouv oi A, B, r, A, E zE,f\z dXXiqXouc; ou [iz- xpouoiv, cpavepov ou8e yap 6 A xov B ^.expeT. Xeyw 8rj, oxi ou8e aXXoc; ou8el<; ou8eva [icxprpei. z\ yap 8u- vaxov, jiexpeixw 6 A xov T. xod oaoi elolv oi A, B, T, xoaouxoi eiXf](p'dwaav IXd/iaxoi dpid^ioi xfiv xov auxov Xoyov e/ovxtov xolc; A, B, T oi Z, H, 0. xod knel oi Z, H, O ev tu auxw X6ya> eiai xou; A, B, T, xai eaxiv laov xo TtXrydoc; x«v A, B, T xw TCXr^ei iwv Z, H, O, 8i' laou dpa eaxiv cbc 6 A 7tp6<; xov T, ouxcoc; 6 Z 7tp6<; xov 8. xod eirei eaxiv cbc; 6 A 7tp6<; xov B, ouxgk 6 Z Ttpoc xov H, oO ^texpel 8e 6 A xov B, ou ^expeT dpa ouSe 6 Z xov H- oux dpa ^tovd<; eaxiv 6 Z- f] yap ^ovac icdvxa dpid^tov ^texpeT. xai eiaiv oi Z, 9 icpwxoi Tipoc; dXXr|Xou<; [ouSe 6 Z dpa xov 9 fiexpel]. xat eaxiv cb<; 6 Z Tipoc; xov 9, oux«<; 6 A jepoe; xov E ou8e 6 A dpa xov T ^texpeT. o^toiog Br] Beic;o^ev, oxi ouSe aXXoc ou8ei<; ouSeva ^texpr]aei- oicep eSei SeT^ai. 'Edv Saiv ouoaoioOv dpiif)[ioi [e?fj<;] dvdXoyov, 6 Be Tipwxog xov ea/axov (jLexpfj, xai xov Seuxepov ^expr]aei. (Which is) the very thing it was required to show. Proposition 6 If there are any multitude whatsoever of continuously proportional numbers, and the first does not measure the second, then no other (number) will measure any other (number) either. A' 1 Bi 1 Ci 1 Di 1 Ei 1 F i 1 G Hi 1 Let A, B, C, D, E be any multitude whatsoever of continuously proportional numbers, and let A not mea- sure B. I say that no other (number) will measure any other (number) either. Now, (it is) clear that A, B, C, D, E do not succes- sively measure one another. For A does not even mea- sure B. So I say that no other (number) will measure any other (number) either. For, if possible, let A measure C. And as many (numbers) as are A, B, C, let so many of the least numbers, F, G, H, have been taken of those (numbers) having the same ratio as A, B, C [Prop. 7.33]. And since F, G, H are in the same ratio as A, B, C, and the multitude of A, B, C is equal to the multitude of F, G, H, thus, via equality, as A is to C, so F (is) to H [Prop. 7.14]. And since as A is to B, so F (is) to G, and A does not measure B, F does not measure G either [Def. 7.20]. Thus, F is not a unit. For a unit measures all numbers. And F and H are prime to one another [Prop. 8.3] [and thus F does not measure H]. And as F is to H, so A (is) to C. And thus A does not measure C either [Def. 7.20]. So, similarly, we can show that no other (number) can measure any other (number) either. (Which is) the very thing it was required to show. Proposition 7 If there are any multitude whatsoever of [continu- ously] proportional numbers, and the first measures the 234 STOIXEIQN rf. ELEMENTS BOOK 8 A' ' B 1 V 1 A> 1 'Eoxcooav onoaoiouv dpid^iol e^fjc; dvdXoyov ol A, B, T, A, 6 Se A xov A [isxperlxw Xeyto, oxi xal 6 A xov B jxexpel. El yap ou (iexpsT 6 A xov B, ou8e dXXoc; oOSeu; ou8eva \L£Tpx]aev jjiexpeT 8e 6 A xov A. fjisxpeT dpa xal 6 A xov B- oTterp e8ei 8eT?ai. *]'• 'Edv 8uo dpii9|ji«v jiexa^u xaxd xo auve)(ec; dvdXoyov e^miKxoaiv dpi%o£, oaoi sic; auxouc; |iexa^u xaxd xo au- v£)(£<; dvoXoyov e^ranxouaiv dpidjiol, xooouxoi xal ek xouc xov auxov Xoyov e^ovxac; [auxolc;] ^£xa<;u xaxd xo ouve)(ec; dvdXoyov 6[i7tEaouvxai A i ' E' ' r ' ' M' ' A ' N' ' B ' ' Z' ' H ' O' ' K 1 ' A' ' Auo yap dpn5^Gv xwv A, B jisxac^u xaxd xo auvexec; dvdXoyov e^mmxexwaav dpid|iol ol T, A, xal 7i£Tioir]o , da) d>c 6 A Ttpoc; xov B, ouxcoc; 6 E npoc; xov Z- Xsyw, oxi oaoi sic, xouc; A, B ^iexac;u xaxd xo auve)(£<; dvdXoyov e^msTtxtoxaaiv dpid^oi, xoaouxoi xal zlc. xoug E, Z ^exac^u xaxd xo auvzyzz dvdXoyov e^Tteaouvxai. "Oaoi ydp rial xcp 7tXrji!tei ol A, B, T, A, xooouxoi riXr|cpiL>Gjaav sXd)(iaxoi dpiiSjiol xfiv xov auxov Xoyov s/ovxiov xoTc; A, r, A, B ol H, 9, K, A- ol dpa axpoi auxwv ol H, A npGxoi upoc dXXf|Xouc; elalv. xal enel ol A, r, A, B xolc; H, 0, K, A ev xa> auxco Xoyio elalv, xal eaxiv I'oov xo TiXfj'doc; xGv A, T, A, B xG txXt^t&si xGv H, 9, K, A, 8i° I'oou dpa eaxlv u;6 A Ttpoc; xov B, ouxgjc; 6 H npoc; xov A. &>c, 8e 6 A npoc; xov B, ouxtog 6 E npoc; xov Z- xal last, then (the first) will also measure the second. A' ' B' 1 C' 1 Di 1 Let A, B, C, D be any number whatsoever of continu- ously proportional numbers. And let A measure D. I say that A also measures B. For if A does not measure B then no other (number) will measure any other (number) either [Prop. 8.6]. But A measures D. Thus, A also measures B. (Which is) the very thing it was required to show. Proposition 8 If between two numbers there fall (some) numbers in continued proportion then, as many numbers as fall in between them in continued proportion, so many (num- bers) will also fall in between (any two numbers) having the same ratio [as them] in continued proportion. A i 1 E i 1 Ci 1 M' 1 D' ' N' ' Bi 1 F ' ' G' ' H' ' K' ' L ' ' For let the numbers, C and D, fall between two num- bers, A and B, in continued proportion, and let it have been contrived (that) as A (is) to B, so E (is) to F. I say that, as many numbers as have fallen in between A and B in continued proportion, so many (numbers) will also fall in between E and F in continued proportion. For as many as A, B, C, D are in multitude, let so many of the least numbers, G, H, K, L, having the same ratio as A, B, C, D, have been taken [Prop. 7.33]. Thus, the outermost of them, G and L, are prime to one another [Prop. 8.3]. And since A, B, C, D are in the same ratio as G, H, K, L, and the multitude of A, B, C, D is equal to the multitude of G, H, K, L, thus, via equality, as A is to B, so G (is) to L [Prop. 7.14]. And as A (is) to B, so 235 STOIXEIQN rf. ELEMENTS BOOK 8 tbc; dpa 6 H Ttpoc; tov A, outmc; 6 E Ttpoc; tov Z. ol 6e H, A TtpfiToi, ol 8e upwxoi xal eXdxioToi, ol 8e eXdxioToi dpi'diiol (iETpouai touc; tov auTov Xoyov exovTac iadxic; 6 tc iieiCiov xov [iei^ova xal 6 eXdaatov tov eXdaaova, TouTeaTiv 6 Te fjyouijievoc; tov fjyouiievov xal 6 eTtojievoc; tov eTiojievov. iadxic; apa 6 H tov E ^iCTpeT xal 6 A tov Z. oadxic 8f] 6 H tov E [Lsxpei, TooauTaxic; xal exaTcpoc; to>v 0, K exaTepov tGv M, N LiETpehto- oi H, 9, K, A apa touc; E, M, N, Z iadxic; [iETpouaiv. oi H, 6, K, A apa tou; E, M, N, Z ev tG auTfi Xoycp eiaiV dXXd ol H, O, K, A toTc; A, T, A, B ev t£> auTW Xoyio eiaiv xal ol A, T, A, B apa toTc; E, M, N, Z ev tw auTcp Xoycp eiaiv. oi 8e A, T, A, B eJ;rjc; dvdXoyov eiaiv xal oi E, M, N, Z apa e^fjg dvdXoyov eiaiv. oaoi apa eic; touc; A, B ^iCTac;u xaTa to auvexec; dvdXoyov eiineiiTtoxaaiv dpnSiioi, ToaouToi xal eic; touc; E, Z [ieTa^u xaTa to auvexec; dvdXoyov eii7ien;T«xaaiv dpid^oi- orcep e8ei 8eTc;ai. E (is) to F. And thus as G (is) to L, so E (is) to F. And G and L (are) prime (to one another). And (numbers) prime (to one another are) also the least (numbers hav- ing the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measur- ing) the greater, and the lesser the lesser — that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, G measures E the same number of times as L (measures) F. So as many times as G measures E, so many times let H, K also measure M, N, respectively. Thus, G, H, K, L measure E, M, N, F (respectively) an equal number of times. Thus, G, H, K, L are in the same ratio as E, M, N, F [Def. 7.20] . But, G, H, K, L are in the same ratio as A, C, D, B. Thus, A, C, D, B are also in the same ratio as E, M, N, F. And A, C, D, B are continuously proportional. Thus, E, M, N, F are also continuously proportional. Thus, as many numbers as have fallen in between A and B in continued proportion, so many numbers have also fallen in between E and F in continued proportion. (Which is) the very thing it was required to show. Proposition 9 'Edv 8uo dpii9|jiol TtpwToi npoc; dXXrjXouc; Saiv, xal eic; auTouc; ^CTac;u xaTa to auvexec; dvdXoyov e^TUKToaiv dprd^toi, oaoi eic; auTouc; ^CTa^u xaTa to auvexec; dvdXoyov eiiTUTiTouaiv dpid^toi, ToaoOroi xal exaTepou auT«v xal iiovd8o<; iieTac^u xaTa to auve/ec; dvdXoyov e^uieaouvTai. A i 1 ©i 1 r i- K> Ah B h Eh O 'EaTwaav 8uo dpid^ol KpWTOi Ttpoc; dXXrjXouc; oi A, B, xal eic; auTouc; ^.eTa^u xaTa to auvexec; dvdXoyov e^TUKTCToaav oi T, A, xal exxeiadw f) E ^tovdc;- Xeyw, oti oaoi eic touc; A, B ^CTac^u xaTa to auvexec; dvdXoyov e^meTiTWxaaiv dpid^ioi, ToaouToi xal exaTepou t«v A, B xal xfjc; jiovdBoc; iieTa^u xaTa to auvexec; dvdXoyov eiiTieaouvTai. EiX^cp-dwaav yap 8uo (iev dpi'diiol eXdxioToi ev t£3 twv A, T, A, B Xoycp ovtcc; oi Z, H, TpeTc; 8e oi <d, K, A, xal del If two numbers are prime to one another and there fall in between them (some) numbers in continued pro- portion then, as many numbers as fall in between them in continued proportion, so many (numbers) will also fall between each of them and a unit in continued proportion. A i 1 H 1 K Dh F i 1 O G 1 P 1 Let A and B be two numbers (which are) prime to one another, and let the (numbers) C and D fall in be- tween them in continued proportion. And let the unit E be set out. I say that, as many numbers as have fallen in between A and B in continued proportion, so many (numbers) will also fall between each of A and B and the unit in continued proportion. For let the least two numbers, F and G, which are in the ratio of A, C, D, B, have been taken [Prop. 7.33]. 236 STOIXEIQN rf. ELEMENTS BOOK 8 sSfjc; evl tiXeiouc;, ecoc; dv taov yevrjxai to TCXrydoc; auxfiiv xfi TtXn^T&Ei xfiv A, T, A, B. siXr]cpi9«oav, xal eaxcoaav ol M, N, S, O. cpavspov 8r), oxi 6 [isv Z sauxov 7toXXaTtXaaidaa<; xov TCSTioirjXEv, xov Ss TroXXaTtXaaidaac; xov M tiettoi/jxev, xal 6 H eauxov (isv rcoXXaTiXaaidaac; xov A TtETtoirjxsv, xov 8e A TtoXXanXaaidaac; xov O TtETtoirjxsv. xal etc! oi M, N, S, O eXd)(iaxo[ siai xtov xov auxov Xoyov e/ovxwv xoT<; Z, H, eial 6s xal oi A, T, A, B sXa/ioxoi xfiv xov auxov Xoyov exovxcov xou; Z, H, xai saxiv laov xo TxXfjiE>oc; xwv M, N, S, O xG TtXr] n &eL xwv A, V, A, B, sxaaxoc; dpa xwv M, N, S, O exdaxw xwv A, T, A, B lao? eaxiv l'ao<; dpa saxlv 6 [isv M iw A, 6 8e O iw B. xal etie! 6 Z eauxov TroXXaTtXaaidaac; xov O KETXoirjxev, 6 Z dpa xov ^lexpei xaxa xa<; ev xw Z ^iovd8a<;. [icxpel 8e xal f\ E ^tova<; xov Z xaxa xd<; sv auxw ^ovd8a<;- ladxic; dpa f] E ^tovdc; xov Z apidjjiov jiexpsT xal 6 Z xov O. eaxiv dpa &>q rj E (iova<; 7tp6<; xov Z dpid^tov, ouxoc 6 Z Kp6<; xov O. ndXiv, end 6 Z xov 9 TtoXXaTtXaaidaac; xov M TienoirjXEv, 6 dpa xov M piexpeT xaxa xdc sv xw Z ^iovd8a<;. jiexpeT 8e xal f] E (lovag xov Z dpid^iov xaxa xa<; ev auxo ^iovd8a<;- ladxic; dpa f) E fiovdc; xov Z dpid^iov ^xexpeT xal 6 xov M. eaxiv dpa w<; f\ E |iova<; 7ip6<; xov Z dpid^iov, ouxcoc; 6 O 7ip6<; xov M. £8ei)fdr] 8e xal cb<; f) E fiovdc; rcpoc; xov Z dpid^iov, ouxw<; 6 Z 7ip6<; xov 8' xal «<; dpa f] E fiovdc; Ttpoc; xov Z dpi-djiov, ouxw<; 6 Z Ttpoc; xov xal 6 icpoc; xov M. iao<; 8e 6 M x£) A - eaxiv dpa (be; f] E fiovdc; Ttpoc; xov Z dpidfiov, ouxwc; 6 Z Ttpoc; xov xal 6 Ttpoc; xov A. 8id xd auxd 8r] xal 6<; f] E [iovac; Ttpoc; xov H dpid^iov, ouxgx 6 H Ttpoc; xov A xal 6 A Ttpoc; xov B. oaoi dpa ei? xouc; A, B fiexac;u xaxa xo auve^ec; dvdXoyov efjuteTtxcoxaaiv dpi-dfioi, xoaouxoi xal exaxepou xwv A, B xal (iovdSoc; xrjc; E fiexac;u xaxa xo auvexec; dvdXoyov e(iTteTtxoxaaiv dpi-dfior OTtep e8ei 8eTc;ai. And the (least) three (numbers), H, K, L. And so on, successively increasing by one, until the multitude of the (least numbers taken) is made equal to the multitude of A, C, D, B [Prop. 8.2]. Let them have been taken, and let them be M, N, O, P. So (it is) clear that F has made H (by) multiplying itself, and has made M (by) multi- plying H . And G has made L (by) multiplying itself, and has made P (by) multiplying L [Prop. 8.2 corr.]. And since M, N, O, P are the least of those (numbers) hav- ing the same ratio as F, G, and A, C, D, B are also the least of those (numbers) having the same ratio as F, G [Prop. 8.2], and the multitude of M, N, O, P is equal to the multitude of A, C, D, B, thus M, N, O, P are equal to A, C, D, B, respectively. Thus, M is equal to A, and P to B. And since F has made H (by) multiply- ing itself, F thus measures H according to the units in F [Def. 7.15]. And the unit E also measures F according to the units in it. Thus, the unit E measures the number F as many times as F (measures) H. Thus, as the unit E is to the number F, so F (is) to H [Def. 7.20]. Again, since F has made M (by) multiplying H, H thus measures M according to the units in F [Def. 7.15]. And the unit E also measures the number F according to the units in it. Thus, the unit E measures the number F as many times as H (measures) M. Thus, as the unit E is to the number F, so H (is) to M [Prop. 7.20]. And it was shown that as the unit E (is) to the number F, so F (is) to H. And thus as the unit E (is) to the number F, so F (is) to H, and H (is) to M. And M (is) equal to A. Thus, as the unit E is to the number F, so F (is) to H, and H to A. And so, for the same (reasons), as the unit E (is) to the number G, so G (is) to L, and L to B. Thus, as many (numbers) as have fallen in between A and B in continued proportion, so many numbers have also fallen between each of A and B and the unit E in continued proportion. (Which is) the very thing it was required to show. i . 'Edv Suo dpid^Gv exaxepou xal ^iovd8o<; jiexa^u xaxa xo auvEXE? dvdXoyov eututixcoctiv dpid^toi, oaoi sxaxspou auxwv xal jiovd8o^ \Lzxa£,b xaxa xo auvE)(E<; dvdXoyov s^TUTtxouaiv dpid^toi, xoaouxoi xal sic; auxouc [isxa$u xaxa xo auvE^EC dvdXoyov E^msaouvxai. Auo yap dpi^wv xwv A, B xal [iovdSoc xfjc T \ie- xa^u xaxa xo auve)(s<; dvdXoyov EjjiTUTixExcoaav &pid|jioi ol xe A, E xal oi Z, H- Xsyw, oxi oaoi exaxspou xwv A, B xal (jiovd8o<; xfjc T [isxa^u xaxa xo auvs)(£<; dvdXoyov E^KETixcoxaaiv dpvd\Lol, xoaouxoi xal zlc, xou<; A, B ^.sxa^u xaxa xo auvE^EC dvdXoyov suTisaouvxai. Proposition 10 If (some) numbers fall between each of two numbers and a unit in continued proportion then, as many (num- bers) as fall between each of the (two numbers) and the unit in continued proportion, so many (numbers) will also fall in between the (two numbers) themselves in con- tinued proportion. For let the numbers D, E and F, G fall between the numbers A and B (respectively) and the unit C in con- tinued proportion. I say that, as many numbers as have fallen between each of A and B and the unit C in contin- ued proportion, so many will also fall in between A and B in continued proportion. 237 STOIXEIQN rf. ELEMENTS BOOK 8 r — ' r — ' A' ' Z ' i: — i ii ' A 1 ' B ' ' ©i i K 1 ' A' ' O A y&P TOV Z TioXXaTiXaaidaac; xov tioicixo, exdxepoc; 8e iwv A, Z xov 6 TioXXaTiXaaidaac; exdxepov XO)V K, A TCOIEITM. Kod ctici eaxiv (be; rj T [iovdc; Tipoc; tov A dpid^iov, ouxox 6 A Tipoc; xov E, ladxic; dpa f] T [lovac; xov A dpid^iov ^.expel xal 6 A xov E. f) 8e r jiovdc; xov A dpid^tov ^.expeT xaxd xdc; ev tu A [iovd8ac xal 6 A dpa dpnf)[i6<; xov E ^xexpeT xaxd xdc; ev xw A ^.ovdSac;' 6 A dpa eauxov TioXXaTiXaaidaac; xov E -rcsTCofyxev. TidXiv, eTiei eaxiv (be; f) T [[lovdc;] Tipoc; xov A dpn)^6v, oux«c; 6 E Tipoc; xov A, ladxic; dpa f] T \±ovaz xov A dpi/d^iov ^.expeT xal 6 E xov A. f) 8e T [iovdc; xov A dpi-djiov ^.expel xaxd xdc; ev xw A ^.ovdBac;- xal 6 E dpa xov A fjiexpeT xaxd xdc; ev xw A [iovd5a<;- 6 A dpa xov E TioXXaTiXaaidaac; xov A 7i£Tio(r]X£v. Bid xd auxd 8r] xal 6 \ie\) Z eauxov TioXXaTiXaaidaac; xov H TiCTio(r]xev, xov 6e H TioXXaTiXaaidaac; xov B Ttenoirjxev. xal excel 6 A eauxov [iev TioXXaTiXaaidaac; xov E TiCTioirjxev, xov Be Z TioXXaTiXaaidaac; xov TteTioirjxev, eaxiv dpa q;6 A Tipoc; xov Z, ouxoc; 6 E Tipoc; xov 0. 8id xd auxd Srj xal (be; 6 A Tipoc; xov Z, ouxwc; 6 Tipoc; xov H. xal (be; dpa 6 E Tipoc; xov 0, ouxmc; 6 G Tipoc; xov H. TidXiv, CTiel 6 A exdxepov xwv E, TioXXaTiXaaidaac; exdxepov xwv A, K TiCTioirjxev, eaxiv dpa w;6E Tipoc; xov 0, ouxcoc; 6 A Tipoc; xov K. dXX' (be; 6 E Tipoc; xov 0, ouxoc; 6 A Tipoc; xov Z - xal (i><; dpa 6 A Tipoc; xov Z, ouxwc; 6 A Tipoc; xov K. TidXiv, ensl exdxepoc xwv A, Z xov TioXXaTiXaaidaac; exdxepov xwv K, A TiCTioirjxev, eaxiv dpa (be; 6 A Tipoc; xov Z, ouxok 6 K Tipoc; xov A. dXX' (be; 6 A Tipoc; xov Z, ouxwc; 6 A Tipoc; xov K- xal (be; dpa 6 A Tipoc; xov K, ouxwc; 6 K Tipoc; xov A. exi eTiel 6 Z exdxepov xwv 0, H TioXXaTiXaaidaac; exdxepov xwv A, B TiCTioirjxev, eaxiv dpa (be; 6 Tipoc; xov H, ouxwc; 6 A Tipoc; xov B. (be; 5e 6 Tipoc; xov H, ouxwc; 6 A Tipoc; xov Z' xal (be; dpa 6 A Tipoc; xov Z, ouxwc; 6 A Tipoc; xov B. e8e[)fdr] 8e xal (be; 6 A Tipoc; xov Z, ouxwc; o xe A Tipoc; xov K xal 6 K Tipoc; xov A' xal (be; dpa 6 A Tipoc; xov K, ouxwc; 6 K Tipoc; xov A xal 6 A Tipoc; xov B. oi A, K, A, B dpa xaxd xo auvexec; e$fjc; eiaiv dvdXoyov. oaoi dpa exaxepou xwv A, B xal xrjc; T ^.ovdBoc; jiexac;u xaxd xo auve^ec; dvdXoyov e^TUTixouaiv dpid^toi, xoaouxoi xal eic; xouc; A, B ^icxa^u xaxd xo auvexec; epmeaoOvxai- oiiep e8ei O — ' C' — ' D F i 1 E i 1 G' 1 A i 1 B i 1 Hi ' K L ' ' For let D make H (by) multiplying F. And let D, F make K, L, respectively, by multiplying H . As since as the unit C is to the number D, so D (is) to E, the unit C thus measures the number D as many times as D (measures) E [Def. 7.20]. And the unit C measures the number D according to the units in D. Thus, the number D also measures E according to the units in D. Thus, D has made E (by) multiplying itself. Again, since as the [unit] C is to the number D, so E (is) to A, the unit C thus measures the number D as many times as E (measures) A [Def. 7.20]. And the unit C measures the number D according to the units in D. Thus, E also mea- sures A according to the units in D. Thus, D has made A (by) multiplying E. And so, for the same (reasons), F has made G (by) multiplying itself, and has made B (by) multiplying G. And since D has made E (by) multiplying itself, and has made H (by) multiplying F, thus as D is to F, so E (is) to H [Prop 7.17]. And so, for the same rea- sons, as D (is) to F, so H (is) to G [Prop. 7.18]. And thus as E (is) to H, so H (is) to G. Again, since D has made A, K (by) multiplying E, H, respectively, thus as E is to H, so A (is) to K [Prop 7.17]. But, as E (is) to H, so D (is) to F. And thus as D (is) to F, so A (is) to K. Again, since D, F have made K, L, respectively, (by) multiply- ing H, thus as D is to F, so K (is) to L [Prop. 7.18]. But, as D (is) to F, so A (is) to K. And thus as A (is) to K, so K (is) to L. Further, since F has made L, B (by) mul- tiplying H, G, respectively, thus as H is to G, so L (is) to B [Prop 7.17]. And as H (is) to G, so D (is) to F. And thus as D (is) to i^, so L (is) to £?. And it was also shown that as D (is) to F, so A (is) to AT, and K to i. And thus as A (is) to K, so if (is) to L, and i to £?. Thus, A, K, L, B are successively in continued proportion. Thus, as many numbers as fall between each of A and B and the unit C in continued proportion, so many will also fall in between A and B in continued proportion. (Which is) the very thing it was required to show 238 STOIXEIQN rf. ELEMENTS BOOK 8 8eT<;ai. ia'. Auo xexpaycbvcov dpnD^aiv zlc, [isaoc; dvdXoyov saxiv dprd^ioc;, xdi 6 xsxpdyiovoc; Ttpoc; xov xexpdywvov 81- TtXamova Xoyov *F e P ^ ^Xsupd Ttpoc; x/]v TtXeupdv. A i ' B i 1 r i 1 a i 1 Ei 1 "Eaxwaav xexpdywvoi dpid^iol oi A, B, xdi xoO [lev A TtXeupd eaxio 6 T, xoO 8s B 6 A- Xeyio, oxi xfiv A, B eTc; [isaoc, dvdXoyov eaxiv dpidjjioc;, xdi 6 A Ttpoc; xov B 8i- TtXamova Xoyov zyz\ fjnep 6 T Ttpoc; xov A. O r yap xov A TtoXXaitXaaidaac; xov E Ttoieixw. xdi etieI xexpdywvoc; eaxiv 6 A, TtXeupd 8e auxou eaxiv 6 T, 6 T dpa eauxov TtoXXaTtXaaidaac; xov A TteTtoirjxev. 8id xd auxd 6r] xdi 6 A eauxov TtoXXaTtXaaidaac; xov B TteTtoirjxev. CTtel ouv 6 r exdxepov xSv T, A TtoXXaTtXaaidaac; exdxepov xSv A, E TteTtoirjxev, eaxiv dpa 6 T Ttpoc; xov A, ouxioc; 6 A Ttpoc; xov E. 6ia xd auxd 8f) xal &>c, 6 T Ttpoc; xov A, ouxioc; 6 E Ttpoc; xov B. xal tbc; dpa 6 A Ttpoc; xov E, ouxcoc; 6 E Ttpoc; xov B. xaiv A, B dpa eTc; jjiaoc; dvdXoyov eaxiv dpi'djioc;. Aeyw 8rj, oxi xal 6 A Ttpoc; xov B BiTtXaaiova Xoyov e)(ei f)Tiep 6 r Ttpoc xov A. ETiel yap xpelc; dpi'djiol dvdXoyov eiaiv ol A, E, B, 6 A dpa Ttpoc; xov B BiTtXaaiova Xoyov eyei f\Kep 6 A Ttpoc; xov E. uc; 8e 6 A Ttpoc; xov E, ouxwc; 6 T Ttpoc; xov A. 6 A dpa Ttpoc; xov B BiTtXaaiova Xoyov e)(ei fjitep f) T TtXeupd Ttpoc; x/]v A- oTtep eSei Belial. Proposition 11 There exists one number in mean proportion to two (given) square numbers.^ And (one) square (number) has to the (other) square (number) a squared^ ratio with respect to (that) the side (of the former has) to the side (of the latter). A' ' B " 1 O 1 D' 1 Ei 1 Let A and B be square numbers, and let C be the side of A, and D (the side) of B. I say that there exists one number in mean proportion to A and B, and that A has to B a squared ratio with respect to (that) C (has) to D. For let C make E (by) multiplying D. And since A is square, and C is its side, C has thus made A (by) multi- plying itself. And so, for the same (reasons), D has made B (by) multiplying itself. Therefore, since C has made A, E (by) multiplying C, D, respectively, thus as C is to D, so A (is) to E [Prop. 7.17]. And so, for the same (rea- sons), as C (is) to D, so E (is) to B [Prop. 7.18]. And thus as A (is) to E, so E (is) to B. Thus, one number (namely, E) is in mean proportion to A and B. So I say that A also has to B a squared ratio with respect to (that) C (has) to D. For since A, E, B are three (continuously) proportional numbers, A thus has to B a squared ratio with respect to (that) A (has) to E [Def. 5.9]. And as A (is) to E, so C (is) to D. Thus, A has to B a squared ratio with respect to (that) side C (has) to (side) D. (Which is) the very thing it was required to show. t In other words, between two given square numbers there exists a number in continued proportion. * Literally, "double". IP'. Auo xupwv dpiduov Buo ^xeaoi dvdXoyov eiaiv dpi%o(, xal 6 xu(3o<; Ttpoc xov xupov xpiTtXaaiova Xoyov ^y^l Wep f\ TtXeupd Ttpoc; xr)v TtXeupdv. TEaxoaav xupoi dpi-dpiol oi A, B xal xou ^iev A TtXeupd eaxco 6 T, xou 8s B 6 A- Xeyw, oxi xGv A, B Buo ^eaoi dvdXoyov eiaiv dpi-djioi, xal 6 A Ttpoc; xov B xpiTtXaaiova Xoyov e/ei fjTtep 6 T Ttpoc; xov A. Proposition 12 There exist two numbers in mean proportion to two (given) cube numbers. t And (one) cube (number) has to the (other) cube (number) a cubed^ ratio with respect to (that) the side (of the former has) to the side (of the latter). Let A and B be cube numbers, and let C be the side of A, and D (the side) of B. I say that there exist two numbers in mean proportion to A and B, and that A has 239 STOIXEIQN rf. ELEMENTS BOOK 8 A i 1 Ei 1 B i 1 Zi 1 r i 1 Hi 1 A i 1 ©i 1 K' ' 'O yap r eauxov [iev TioXXauXaaidaac; xov E tioicixgj, xov 8e A TtoXXauXaaidaac; tov Z Tioieixo, 6 8e A eauxov TtoXXauXaaidaac; xov H Ttoieixw, exdxepoc; 8e xwv T, A xov Z TtoXXaTtXaaidaac; exdxepov xwv 0, K Ttoieixo. Kod etxsl xu|3oc; eaxiv 6 A, TiXeupa 8e auxou 6 T, xal 6 r srauxov ^.ev TtoXXauXaaidaac; xov E Ttcnoirjxev, 6 T apa eauxov (iev TtoXXaTtXaaidaac; xov E TteTtoir]xev, xov 8e E TtoXXaTiXaaidaac; xov A TteTiofyxev. Bid xd auxa Sr] xal 6 A eauxov ^iev TtoXXaTtXaaidaac; xov H TteTtoir)xev, xov 8e H TtoXXaTtXaaidaac; xov B TteTtoirjxev. xal eitei 6 T exdxepov iwv T, A TtoXXaTtXaaidaac; exdxepov xwv E, Z TteTtoirjxev, eaxiv apa (be; 6 T Ttpoc; xov A, ouxwc; 6 E Ttpoc; xov Z. 8id xd auxa 8f) xal (be; 6 T Ttpoc; xov A, ouxwc; 6 Z Ttpoc; xov H. TtdXiv, euel 6 T exdxepov xQv E, Z TtoXXaTtXaaidaac; exdxepov xQv A, TteTtoir]xev, eaxiv apa (be; 6 E Ttpoc; xov Z, oux«<; 6 A Ttpoc; xov 6. (be; 8e 6 E Ttpoc; xov Z, ouxwc; 6 T Ttpoc; xov A- xal (be; apa 6 T Ttpoc; xov A, ouxwc; 6 A Ttpoc; xov 0. TtdXiv, CTtel exdxepoc; xwv T, A xov Z TtoXXaTtXaaidaac; exdxepov xwv 0, K TteTtoirjxev, eaxiv apa u?6T Ttpoc; xov A, ouxwc; 6 Ttpoc; xov K. TtdXiv, CTtel 6 A exdxepov iSv Z, H TtoXXaTtXaaidaac; exdxepov xwv K, B TteTtoirjxev, eaxiv apa (be 6 Z Ttpoc; xov H, ouxw<; 6 K Ttpoc; xov B. cLk 8e 6 Z Ttpoc; xov H, ouxwc; 6 T Ttpoc; xov A- xal (be; apa 6 T Ttpoc; xov A, ouxwc; o xe A Tipoc; xov xal 6 Ttpoc; xov K xal 6 K Tipoc; xov B. xebv A, B apa 8uo ueaoi dvdXoyov eiaiv oi 0, K. Aeyw 8rj, oxi xal 6 A Ttpoc; xov B xpiTtXaaiova Xoyov e^ei r]Ttep 6 r Tipoc; xov A. euel yap xeaaapec; dpid^iol dvdXoyov eiaiv oi A, 0, K, B, 6 A apa Tipoc; xov B xpiTtXaaiova Xoyov e^ei f)Tiep 6 A Tipoc; xov 0. (be; Se 6 A Tipoc; xov 0, ouxoc; 6 T Tipoc; xov A- xal 6 A [apa] Tipoc; xov B xpiTtXaaiova Xoyov exei fjnep 6 T Tipoc; xov A- oTtep e8ei 8eTc;ai. t In other words, between two given cube numbers there exist two nun t Literally, "triple". to B a cubed ratio with respect to (that) C (has) to D. A i 1 E i 1 B i 1 F i 1 Ci 1 Gi 1 Di 1 Hi 1 Ki 1 For let C make E (by) multiplying itself, and let it make F (by) multiplying D. And let D make G (by) mul- tiplying itself, and let C, D make H, K, respectively, (by) multiplying F. And since A is cube, and C (is) its side, and C has made E (by) multiplying itself, C has thus made E (by) multiplying itself, and has made A (by) multiplying E. And so, for the same (reasons), D has made G (by) mul- tiplying itself, and has made B (by) multiplying G. And since C has made E, F (by) multiplying C, D, respec- tively, thus as C is to D, so E (is) to F [Prop. 7.17]. And so, for the same (reasons), as C (is) to D, so F (is) to G [Prop. 7.18]. Again, since C has made A, H (by) multi- plying E, F, respectively, thus as E is to F, so A (is) to H [Prop. 7.17]. And as E (is) to F, so C (is) to D. And thus as C (is) to D, so A (is) to H. Again, since C, D have made iJ, K, respectively, (by) multiplying F, thus as C is to D, so _ff (is) to K [Prop. 7.18]. Again, since D has made K, B (by) multiplying F, G, respectively, thus as F is to G, so K (is) to B [Prop. 7.17]. And as F (is) to G, so C (is) to £>. And thus as C (is) to D, so A (is) to H, and to i\T, and K to B. Thus, iJ and K are two (numbers) in mean proportion to A and B. So I say that A also has to B a cubed ratio with re- spect to (that) C (has) to D. For since A, H, K, B are four (continuously) proportional numbers, A thus has to B a cubed ratio with respect to (that) A (has) to H [Def. 5.10]. And as A (is) to H, so C (is) to D. And [thus] A has to B a cubed ratio with respect to (that) C (has) to D. (Which is) the very thing it was required to show. in continued proportion. 'Edv fiaiv oaoiSrjTioxouv dpid^tol e^fjc; dvdXoyov, xal TioXXaTiXaaidaac; exaaxoc eauxov Tioi/j xiva, oi yevojievoi ec; auxwv dvdXoyov eaovxai' xal edv oi ec; dp^fjc xouc yevo^ievouc; TtoXXauXaaidaavxec; Ttoiwai xivac;, xal auxol dvdXoyov eaovxai [xal del Ttepl xouc; axpouc; xouxo au^paivei] . 'Eaxwaav oTioaoiouv dpid^iol ec^fjc; dvdXoyov, oi A, B, Proposition 13 If there are any multitude whatsoever of continuously proportional numbers, and each makes some (number by) multiplying itself, then the (numbers) created from them will (also) be (continuously) proportional. And if the original (numbers) make some (more numbers by) multiplying the created (numbers) then these will also 240 STOIXEIQN rf. ELEMENTS BOOK 8 r, q; 6 A Ttpoc; xov B, ouxioc; 6 B 7tp6<; xov T, xod oi A, B, r eauxouc; [iev TToXXaitXaaLaaavxsc; xoug A, E, Z noieixtoaav, xou<; Se A, E, Z TtoXXaTtXaaidaavxec; xou<; H, 9, K Tioisraoaav Xeyw, ° Tl of Te A, E, Z xal oi H, 9, K zZjf\Z dvdXoyov sioiv. A B r M N O n A E Z H K ^iev yap A xov B TCoXXanXamdaac; xov A toieixco, sxdxspog 8e xwv A, B xov A 7ioXXaTtXaaidaa<; exdxepov iSv M, N Tioieixw. xal TtdXiv 6 \±kv B xov T noXXajiXaaidaa^ xov S Ttoisixio, sxdxspog 8s xfiv B, T xov S TTOXXanXaaidaac; SXaXSpOV XWV O, II 7IOIEIX63. 'O^ioicdc; Sr| idle, erndvco Sel^ojiev, oxi oi A, A, E xal oi H, M, N, 9 'tE,f\z eiaiv dvdXoyov ev xw xou A 7tp6<; xov B Xoyw, xal exi oi E, S, Z xai oi 9, O, II, K zE,f\c, sioiv dvdXoyov sv ifi xou B upoc xov T Xoyco. xai eaxiv w<; 6 A npoc; xov B, ouxck 6 B npbc, xov E xal oi A, A, E apa xolc; E, S, Z ev xo auxo Xoyw elal xal exi oi H, M, N, 9 xolc 9, O, II, K. xai eaxiv laov xo [Lev xwv A, A, E TtXrji9o<; xw xov E, S, Z TiXri'dei, xo 8s xov H, M, N, 9 xw xGv 9, O, II, K' Si' i'aou apa eaxiv <b<; ^xev 6 A npoc; xov E, ouxwc; 6 E 7ip6<; xov Z, (b? 8e 6 H npoc xov 9, ouxgk 6 9 Ttpoc; xov K- oTiep eSei 8eic;ai. be (continuously) proportional [and this always happens with the extremes] . Let A, B, C be any multitude whatsoever of contin- uously proportional numbers, (such that) as A (is) to B, so B (is) to C. And let A, B, C make D, E, F (by) multiplying themselves, and let them make G, H, K (by) multiplying D, E, F. I say that D, E, F and G, H, K are continuously proportional. A' ' L O F >- O M N P Q H>- K For let A make L (by) multiplying B. And let A, B make M, N, respectively, (by) multiplying L. And, again, let B make O (by) multiplying C. And let B, C make P, Q, respectively, (by) multplying O. So, similarly to the above, we can show that D, L, E and G, M, N, H are continuously proportional in the ratio of A to B, and, further, (that) E, O, F and H, P, Q, K are continuously proportional in the ratio of B to C. And as A is to B, so B (is) to C. And thus D, L, E are in the same ratio as E, O, F, and, further, G, M, N, H (are in the same ratio) as H, P, Q, K. And the multitude of D, L, E is equal to the multitude of E, O, F, and that of G, M, N, H to that of H, P, Q, K. Thus, via equality, as D is to E, so E (is) to F, and as G (is) to H, so H (is) to K [Prop. 7.14]. (Which is) the very thing it was required to show. 18'. °Edv xexpdywvoc; xsxpdyovov ^expfj, xai f) nXsupd xr)v TtXeupdv [isxprpei- xal eav f] TtXeupa xrjv nXeupdv (iexpfj, xal 6 xexpdywvoc; xov xexpdywvov [letprpei. 'Eaxwaav xexpdywvoi dpid^iol oi A, B, nXeupai 8e auxwv Eaxcoaav oi T, A, 6 8s A xov B ^sxpdxco - Xeyto, oxi xai 6 T xov A ^expsT. Proposition 14 If a square (number) measures a(nother) square (number) then the side (of the former) will also mea- sure the side (of the latter) . And if the side (of a square number) measures the side (of another square number) then the (former) square (number) will also measure the (latter) square (number). Let A and B be square numbers, and let C and D be their sides (respectively). And let A measure B. I say that C also measures D. 241 STOIXEIQN rf. ELEMENTS BOOK 8 a i 1 r i 1 B i ' A i 1 Ei ' "O r yap tov A TioXXauXaaidaac; tov E raDidTW oi A, E, B dpa l^fje; dvdXoyov slaiv sv xG xou T Ttpoc; tov A Xoyco. xal £7td 01 A, E, B s£rj<; dvdXoyov eiaiv, xal ^expsl 6 A tov B, ^.STpsT dpa xal 6 A tov E. xa[ sgtiv cb<; 6 A 7tp6<; tov E, 0UT6X 6 T npbc, tov A- UETpsI dpa xal 6 T tov A. ITdXiv 8r] 6 T tov A ^STpsrcM- Xsyw, oti xal 6 A tov B ^tSTpsT. Tt5v yap auTGv xaTaaxsuaaiJevTOv ouoick 8ei£o^ev, oti oi A, E, B e^f]z dvdXoyov eiaiv £v iu tou T 7ip6<; tov A Xoyco. xal etcsl eaTiv &>c, 6 T npbz tov A, outoc 6 A 7ip6<; tov E, ^STpeT 8e 6 T tov A, ^STpeT dpa xal 6 A tov E. xal eiaiv ol A, E, B i^f]z dvdXoyov ^xeTpeT dpa xal 6 A tov B. 'Edv dpa TSTpdycovot; TSTpdywvov ^CTpfj, xal f] TtXeupa t/]v TtXeupdv [LSTpfioei- xal edv f) nXeupd ttjv nXeupdv (j.expf], xal 6 TSTpdywvoc; tov TSTpdywvov [Lsjpfiaei- oTtep eBei 8eTc;ai. A 1 ' O ' B 1 1 D 1 Ei 1 For let C make E (by) multiplying D. Thus, A, E, B are continuously proportional in the ratio of C to D [Prop. 8.11]. And since A, E, B are continuously pro- portional, and A measures B, A thus also measures E [Prop. 8.7]. And as A is to E, so C (is) to D. Thus, C also measures D [Def. 7.20] . So, again, let C measure D. I say that A also measures B. For similarly, with the same construction, we can show that A, E, B are continuously proportional in the ratio of C to D. And since as C is to D, so A (is) to E, and C measures D, A thus also measures E [Def. 7.20]. And A, E, B are continuously proportional. Thus, A also measures B. Thus, if a square (number) measures a(nother) square (number) then the side (of the former) will also measure the side (of the latter) . And if the side (of a square num- ber) measures the side (of another square number) then the (former) square (number) will also measure the (lat- ter) square (number). (Which is) the very thing it was required to show. is'. 'Edv xu(3oc; dpiduoc; xupov dpid^tov [iSTpr], xal f) TtXeupa t/]v TiXsupdv \±sjpfioei- xal edv f) TtXeupa tt]v icXeupdv ^xexpf), xal 6 xupo<; tov xupov ^iCTprjaei. Kupog yap dpi%6c; 6 A xupov tov B ^CTpeiTM, xal tou y.sv A uXeupa eaTto 6 T, tou 8s B 6 A- Xeyco, oti 6 T tov A [iETpa. a 1 1 r 1 1 B 1 1 A 1 1 E 1 1 Bi ' Hi ' KJ 1 Z 1 'O r yap eauTov TtoXXa7tXaaidaa<; tov E tioicitco, 6 8s A Proposition 15 If a cube number measures a(nother) cube number then the side (of the former) will also measure the side (of the latter) . And if the side (of a cube number) mea- sures the side (of another cube number) then the (for- mer) cube (number) will also measure the (latter) cube (number). For let the cube number A measure the cube (num- ber) B, and let C be the side of A, and D (the side) of B. I say that C measures D. A 1 1 C' 1 B 1 1 D 1 Ei 1 Hi 1 Gi 1 Ki 1 F 1 1 For let C make E (by) multiplying itself. And let 242 STOIXEIQN rf. ELEMENTS BOOK 8 eauxov TCoXXaTtXaaidaac; tov H ttoieixm, xal exi 6 T xov A TtoXXaitXaaidaag xov Z [ttoieixw], exdxepog 8e xaiv T, A xov Z TioAXamXaaidaac; exdxepov xwv 9, K Ttoidxw. cpavspov 8r], oxi ol E, Z, H xod oi A, 0, K, B £^fj<; dvdXoyov eiaiv sv xG xou r Ttpoc; xov A Xoyw. xal etxel oi A, 9, K, B e^fjc; dvdXoyov eiaiv, xal ^texpeT 6 A xov B, ^icxpeT dpa xal xov 9. xai eaxiv cbc 6 A Ttpoc; xov 9, oux«<; 6 T 7tp6<; xov A- ^texpeT dpa xal 6 T xov A. AXXa 8rj ^lexpeixo 6 T xov A- Xeyw, oxi xal 6 A xov B ^exprpei. Twv yap auxwv xaxaaxeuaadevxwv b\±oic)q 8f) 8ei^o^.ev, oxi oi A, 9, K, B e^fjc; dvdXoyov eiaiv ev xw xou V npoc, xov A X6ya>. xal etcei 6 T xov A ^expert, xai eaxiv (i><; 6 T 7ip6<; xov A, ouxw<; 6 A Ttpoc xov 9, xal 6 A dpa xov 9 ^lexper waxe xal xov B ^icxpel 6 A- oicep e8ei BeT^ai. 'Eav xexpdywvoc; dprd^ioc; xexpdywvov dpii9|ji6v y.r) (lexprj, ou8e f\ TtXeupd xr)v TtXeupdv |iexpr|aef xdv f) TtXeupd x/]v TiXeupdv \±r\ [icxpfj, ou8e 6 xexpdywvoc; xov xexpdyovov ^iexpf|aei. a i 1 r i 1 B ' 1 A i 1 'Tiaxwaav xexpdywvoi dpid^ioi oi A, B, nXeupal 8e auxfiiv eaxwaav oi T, A, xal [ir] ^texpeixM 6 A xov B- Xeyw, oxi ou8e 6 T xov A (iexpa. Ei yap \±£Tpei 6 T xov A, \LSTpr\oei xal 6 A xov B. ou ^expel 8e 6 A xov B- ouSe dpa 6 T xov A ^expr|aei. Mr] ^texpeixM [Srj] TtdXiv 6 T xov A- Xeyw, oxi ouSe 6 A xov B [LSTpfioei. Ei yap ^texpeT 6 A xov B, ^iexpf|aei xal 6 T xov A. ou ^texpeT 8e 6 T xov A- oOB' dpa 6 A xov B ^expfjaa- oitep e8el SeTc;ai. 'Eav xupog dpid^icx; xupov dpid^iov y.r] ^exp/j, ou8s f) TiXeupd xrjv TtXeupdv [lexpiqaei- xdv f) TtXeupd xrjv TtXeupdv [ir] tisxpf), ouSe 6 xupoc; xov xu[3ov ^expiqaei. D make G (by) multiplying itself. And, further, [let] C [make] F (by) multiplying D, and let C, D make iJ, if, respectively, (by) multiplying F. So it is clear that E, F, G and A, H, K, B are continuously proportional in the ratio of C to D [Prop. 8.12]. And since A, H, K, B are continuously proportional, and A measures B, (A) thus also measures H [Prop. 8.7]. And as A is to H, so C (is) to D. Thus, C also measures D [Dei. 7.20]. And so let C measure D. I say that A will also mea- sure B. For similarly, with the same construction, we can show that A, H, K, B are continuously proportional in the ratio of C to D. And since C measures D, and as C is to D, so A (is) to H, A thus also measures H [Dei. 7.20]. Hence, A also measures B. (Which is) the very thing it was required to show. Proposition 16 If a square number does not measure a(nother) square number then the side (of the former) will not measure the side (of the latter) either. And if the side (of a square number) does not measure the side (of another square number) then the (former) square (number) will not measure the (latter) square (number) either. A i ' ' C' ' B i 1 Di 1 Let A and B be square numbers, and let C and D be their sides (respectively). And let A not measure B. I say that C does not measure D either. For if C measures D then A will also measure B [Prop. 8.14]. And A does not measure B. Thus, C will not measure D either. [So], again, let C not measure D. I say that A will not measure B either. For if A measures B then C will also measure D [Prop. 8.14]. And C does not measure D. Thus, A will not measure B either. (Which is) the very thing it was required to show. Proposition 17 If a cube number does not measure a(nother) cube number then the side (of the former) will not measure the side (of the latter) either. And if the side (of a cube num- ber) does not measure the side (of another cube number) then the (former) cube (number) will not measure the (latter) cube (number) either. 243 STOIXEIQN rf. ELEMENTS BOOK 8 Ai 1 r> 1 Ai 1 O B 1 Ai 1 Bi 1 D^ Kupog yap apvd\ibc, 6 A xupov dpidjiov xov B [if\ (ie- xpeixo, xoti xou [iev A TtXeupd eaxw 6 T, xou Be B 6 A- Xeyw, oxi 6 T xov A ou ^ETpVjaei. Ei yap [xexpeT 6 T xov A, xal 6 A xov B [iexpf|aei. ou [xexpeT Be 6 A xov B- ouB' dpa 6 T xov A [xexpeT. AXXd 8rj [if] uxxpeixw 6 T xov A- Xeyw, oxi ouBe 6 A xov B [iexpf|aei. Et yap o A xov B (jiexpeT, xal 6 T xov A [istprpei. ou [lejpsi 8e 6 T xov A- ou8' apa 6 A xov B {izjpfioei- oTtep e8ei 8eTe;ai. ir]'. Auo ojioicov eTtiTte8«v dpidjifiv elc; [izaoz dvdXoyov eaxiv dprd^ioc;- xal 6 enimboz Ttpoc; xov emTteSov BiTtXaaiova Xoyov exei f] by-oXoyoc, TtXeupd Ttpoc; xf)v o^ioXoyov TtXeupdv. A i 1 B i - 1 T i 1 Ei 1 A i 1 Z i 1 For let the cube number A not measure the cube num- ber B. And let G be the side of A, and D (the side) of B. I say that G will not measure D. For if G measures D then A will also measure B [Prop. 8.15]. And A does not measure B. Thus, G does not measure D either. And so let G not measure D. I say that A will not measure B either. For if A measures B then G will also measure D [Prop. 8.15]. And G does not measure D. Thus, A will not measure B either. (Which is) the very thing it was required to show. Proposition 18 There exists one number in mean proportion to two similar plane numbers. And (one) plane (number) has to the (other) plane (number) a squared^ ratio with respect to (that) a corresponding side (of the former has) to a corresponding side (of the latter). A i 1 B i 1 Ci 1 E i— ' Di F i H 1 Tiaxwaav Buo o^ioioi CTtiTteSoi dpid^tol oi A, B, xal xou [lev A TtXeupal eaxioaav oi T, A dpidjioi, xou Be B oi E, Z. xal CTtel o^ioioi CTtiTteBoi eiaiv oi dvdXoyov e^ovxec; xdc; TtXeupdc;, eaxiv apa wq 6 T Ttpoc; xov A, ouxwc; 6 E Ttpoc; xov Z. Xeyco ouv, oxi xwv A, B etc; \izaoc, dvdXoyov eaxiv dpii9|ji6c, xal 6 A Ttpoc; xov B BiTtXaaiova Xoyov e/ei f)Ttep 6 T Ttpoc; xov E f) 6 A Ttpoc; xov Z, xouxeaxiv rjnep f\ o^ioXoyoc; TtXeupd Ttpoc; xr)v o^ioXoyov [TtXeupdv]. Kal eitel eaxiv tbc; 6 T Ttpoc; xov A, ouxcog 6 E Ttpoc; xov Z, evaXXdi; apa eaxiv &>c, 6 T Ttpoc; xov E, 6 A Ttpoc; xov Z. xal eitel sitiiteSoc; eaxiv 6 A, TtXeupal Be auxou oi T, A, 6 A apa xov r TtoXXaTtXaaidaac; xov A TteTtoirjxev. Bid xd auxa 8r) xal 6 E xov Z TtoXXaTtXaaidaac; xov B TteTtoirjxev. 6 A 8r) xov E TtoXXaTtXaaidaac; xov H Ttoieixw. xal eicel 6 A xov \±zv T TtoXXaTtXaaidaac xov A Tteitofyxev, xov Be E TtoXXa- TtXaaidaac; xov H TtCTtoirjxev, eaxiv apa w<; 6 T Ttpoc; xov E, ouxcoc; 6 A Ttpoc; xov H. dXX' (be; 6 T itpoc; xov E, [ouxioc;] 6 A Ttpoc; xov Z- xal wc; apa 6 A Ttpoc; xov Z, ouxwc; 6 A Ttpoc; xov H. itdXiv, CTtel 6 E xov [lev A TtoXXaTtXaaidaac; xov H TteTto[/]xev, xov 8e Z TtoXXaTtXaaidaac; xov B TtCTtoirjxev, eaxiv apa u; 6 A Ttpoc; xov Z, ouxwc; 6 H Ttpoc; xov B. e8eix , dr) Be xal u; 6 A Ttpoc; xov Z, ouxwc; 6 A Ttpoc; xov Gi 1 Let A and £? be two similar plane numbers. And let the numbers C, D be the sides of A, and E, F (the sides) of B. And since similar numbers are those having pro- portional sides [Def. 7.21], thus as C is to D, so E (is) to F. Therefore, I say that there exists one number in mean proportion to A and B, and that A has to B a squared ratio with respect to that C (has) to E, or D to F — that is to say, with respect to (that) a corresponding side (has) to a corresponding [side] . For since as C is to D, so E (is) to F, thus, alternately, as C is to E, so D (is) to F [Prop. 7.13]. And since A is plane, and C, D its sides, D has thus made A (by) mul- tiplying C. And so, for the same (reasons), E has made B (by) multiplying F. So let D make G (by) multiplying E. And since £> has made A (by) multiplying C, and has made G (by) multiplying E, thus as C is to E, so A (is) to G [Prop. 7.17]. But as C (is) to E, [so] L> (is) to F. And thus as D (is) to F, so A (is) to G. Again, since E has made G (by) multiplying D, and has made B (by) multi- plying F, thus as D is to F, so G (is) to B [Prop. 7.17]. And it was also shown that as D (is) to F, so A (is) to G. And thus as A (is) to G, so G (is) to B. Thus, A, G, i? are 244 STOIXEIQN rf. ELEMENTS BOOK 8 EL xai cb? dpa 6 A npoc; xov H, ouxgk 6 H npoc; xov B. oi A, H, B apa zEjff, dvdXoyov eiaiv. xc5v A, B dpa dc; [ieaoc; dvdXoyov eaxiv dpi-djioc- Aeyco 8r], oxi xai 6 A npoc; xov B 8mXaaiova Xoyov exei f]nep f] b\±6koxoq nXeupa npoc; xrjv ojaoXoyov nXeupdv, xouxeaxiv r]nep 6 T npoc; xov Erjo A npoc; xov Z. etc! yap oi A, H, B E^fjc; dvdXoyov slaw, 6 A npoc; xov B SmXaaiova Xoyov £)(£i rjnep npoc; xov H. xai eaxiv wc; 6 A npoc; xov H, ouxox 6 xe F npoc; xov E xai 6 A npoc; xov Z. xai 6 A apa npoc; xov B SmXaaiova Xoyov i)(£i rjnep 6 T npoc; xov E rj 6 A npoc; xov Z- onep e8ei Sdc^ai. t Literally, "double". continously proportional. Thus, there exists one number (namely, G) in mean proportion to A and _B. So I say that A also has to B a squared ratio with respect to (that) a corresponding side (has) to a corre- sponding side — that is to say, with respect to (that) G (has) to E, or D to F. For since A, G, B are continuously proportional, A has to B a squared ratio with respect to (that A has) to G [Prop. 5.9]. And as A is to G, so C (is) to E, and Z) to F. And thus A has to B a squared ratio with respect to (that) C (has) to E, or D to F. (Which is) the very thing it was required to show. 10'. Auo opioiov axepewv dpid^icov 8uo [Leaoi dvdXoyov ejjminxouaiv dpid^ioi- xai 6 axepeoc; npoc; xov o^toiov axepeov xpmXaaiova Xoyov eyei f\Ksp f] ojioXoyoc; nXeupa npoc; xrjv o^toXoyov nXeupdv. A i — — ^ — i r ' A^ ' E' ' B i 1 Z i 1 @i 1 K 1 -h M i 1 Ni 1 A 1 1 3 1 1 "Eaxcoaav Suo o^ioioi axepeoi oi A, B, xai xou \xev A nXeupai eaxtoaav oi T, A, E, xou 8s B oi Z, H, 0. xai end S^ioioi axepeoi eiaiv oi dvdXoyov exovxeg xdc; nXeupdc;, eaxiv dpa <b<; [lev 6 T npoc; xov A, ouxwc; 6 Z npoc; xov H, «<; Be 6 A npoc; xov E, ouxwc; 6 H npoc; xov 9. Xeyw, oxi iSv A, B Suo \ieaoi dvdXoyov ejininxouaiv dptd^toi, xai 6 A npoc; xov B xpmXaaiova Xoyov e)(ei fjnep 6 T npoc; xov Z xai 6 A npoc; xov H xai exi 6 E npoc; xov 0. c O r yap xov A noXXanXaaidaac; xov K noieixw, 6 8e Z xov H noXXanXaaidaac; xov A noieixw. xai end oi T, A xoic; Z, H ev x5 auxfi Xoycp eiaiv, xai ex [ie\ x«v T, A eaxiv 6 K, ex 8e xGv Z, H 6 A, oi K, A [dpa] ojioioi enmeBoi eiaiv dpidjioi- x«v K, A dpa etc; ^icaoc; dvdXoyov eaxiv dprd^ioc;. eax« 6 M. 6 M dpa eaxiv 6 ex x«v A, Z, tbc; ev xfi npo xouxou -dewpiqijiaxi eSei/'dr]. xai end 6 A xov [iev T noXXanXaaidaac; xov K nenoi/jxev, xov 8e Z noXXanXaaidaac; xov M nenoi/jxev, eaxiv dpa cbc; 6 T npoc; xov Z, ouxgjc; 6 K npoc; xov M. dXX' o; 6 K npoc; xov M, 6 M npoc; xov A. oi K, M, A dpa ec^rjc; eiaiv dvdXoyov ev Proposition 19 Two numbers fall (between) two similar solid num- bers in mean proportion. And a solid (number) has to a similar solid (number) a cubed^ ratio with respect to (that) a corresponding side (has) to a corresponding side. A i 1 C 1 D Ei 1 B ' ' F ' ' G ' Hi ' K ' Mi 1 Ni 1 L i 1 Oi ' Let A and B be two similar solid numbers, and let G, D, E be the sides of A, and F, G, H (the sides) of B. And since similar solid (numbers) are those having proportional sides [Def. 7.21], thus as C is to D, so F (is) to G, and as D (is) to E, so G (is) to H . I say that two numbers fall (between) A and B in mean proportion, and (that) A has to B a cubed ratio with respect to (that) C (has) to F, and D to G, and, further, E to H. For let G make K (by) multiplying D, and let F make L (by) multiplying G. And since G, D are in the same ratio as F, G, and K is the (number created) from (mul- tiplying) G, D, and L the (number created) from (multi- plying) F, G, [thus] K and L are similar plane numbers [Def. 7.21]. Thus, there exits one number in mean pro- portion to K and L [Prop. 8.18]. Let it be M. Thus, M is the (number created) from (multiplying) D, F, as shown in the theorem before this (one) . And since D has made K (by) multiplying G, and has made M (by) multiplying F, thus as G is to F, so K (is) to M [Prop. 7.17]. But, as 245 STOIXEIQN rf. ELEMENTS BOOK 8 xco xou r Tipoc xov Z Xoyio. xal CTiei eaxiv &>z b T Tipoc xov A, ouxtoc 6 Z Tipoc xov H, evaXXdc' dpa eaxiv cbc 6 T Tipoc xov Z, ouxwc 6 A Tipoc xov H. Bid xa auxa Bf] xal «c 6 A Tipoc xov H, ouxcoc 6 E Tipoc xov 0. oi K, M, A apa e<;rjc etcriv dvdXoyov ev xe xw xou T Tipoc xov Z Xoyw xal tu xoO A Tipoc xov H xal exi xfi xou E Tipoc xov 9. exaxepoc 8r) xov E, O xov M TioXXaTiXaaidaac exdxepov xwv N, 5 itoidxco. xal inel axepeoc eaxiv 6 A, TiXeupal Be auxou eiaiv oi r, A, E, 6 E dpa xov ex xwv T, A TioXXaTiXaaidaac xov A TieTtoi7)xev. 6 Be ex xwv T, A eaxiv 6 K - 6 E dpa xov K TioXXaTiXaaidaac xov A 7ieuo(r]xev. Bid xa auxa Br] xal 6 9 xov A TioXXaTiXaaidaac xov B TieTio[r|Xev. xal eitel 6 E xov K TioXXaTiXaaidaac xov A TieTioirjXev, dXXd ^xfjv xal xov M TioXXaTiXaaidaac xov N TieTioirjXev, eaxiv dpa 6; o K Tipoc xov M, ouxwc 6 A Tipoc xov N. cbc Be 6 K Tipoc xov M, ouxcoc o xe r Tipoc xov Z xal 6 A Tipoc xov H xal exi 6 E Tipoc xov 0- xal cbc apa 6 T Tipoc xov Z xal 6 A Tipoc xov H xal 6 E Tipoc xov 6, ouxcoc 6 A Tipoc xov N. TidXiv, etc el exaxepoc xcov E, xov M TioXXaTiXaaidaac exdxepov xcov N, 5 7ien;oir]xev, eaxiv apa 6 E Tipoc xov 0, ouxcoc 6 N Tipoc xov S. dXX' cbc 6 E Tipoc xov 0, ouxcoc o xe T Tipoc xov Z xal 6 A Tipoc xov IE xal cbc dpa 6 T Tipoc xov Z xal 6 A Tipoc xov H xal 6 E Tipoc xov 0, ouxcoc 6 xe A Tipoc xov N xal 6 N Tipoc xov S. TidXiv, ercel 6 xov M TioXXaTiXaaidaac xov 5 TieTioirjxev, dXXd \ir\v xal xov A TioXXaTiXaaidaac xov B TiCTioirjxev, eaxiv dpa cbc 6 M Tipoc xov A, ouxcoc 6 S Tipoc xov B. dXX' d>c 6 M Tipoc xov A, ouxcoc o xe T Tipoc xov Z xal 6 A Tipoc xov H xal 6 E Tipoc xov 0. xal cbc dpa 6 T Tipoc xov Z xal 6 A Tipoc xov H xal 6 E Tipoc xov 0, ouxcoc ou (iovov 6 5 Tipoc xov B, dXXd xal 6 A Tipoc xov N xal 6 N Tipoc xov S. oi A, N, S, B apa e^fjc eiaiv dvdXoyov ev xolc eipr)[ievoic xcov TiXeupcbv Xoyoic. Aeyco, 6xi xal 6 A Tipoc xov B xpiTiXaaiova Xoyov e/ei fjTtep f\ o^ioXoyoc TiXeupa Tipoc xfjv b\±6koyov TiXeupdv, xouxeaxiv rjTiep 6 r dpid^ioc Tipoc xov Z rj 6 A Tipoc xov H xal exi 6 E Tipoc xov 0. etieI yap xeaaapec dtpi'djjtol e^c dvdXoyov eiaiv oi A, N, S, B, 6 A apa Tipoc xov B xpi- TiXaaiova Xoyov sjsi fpzzp 6 A Tipoc xov N. dXX'' cbc 6 A Tipoc xov N, ouxwc eSeix^ 6 xe T Tipoc xov Z xal 6 A Tipoc xov H xal exi 6 E Tipoc xov 0. xal 6 A dpa Tipoc xov B xpiTiXaaiova Xoyov zyzi fjTiep f\ ojioXoyoc TiXeupa Tipoc xr)V o^ioXoyov TiXeupdv, xouxeaxiv fjTiep 6 T dpid^ioc Tipoc xov Z xal 6 A Tipoc xov H xal exi 6 E Tipoc xov 0- oTiep eBei BeT^ai. t Literally, "triple". X . 'Edv Buo dpidfjicbv elc ^xeaoc dvdXoyov E^uuTtxfj dpnf)^6c, ojjioioi ctiiticBoi saovxai oi dtpi'djiof. K (is) to M, (so) M (is) to L. Thus, K, M, L are contin- uously proportional in the ratio of C to F. And since as C is to D, so F (is) to G, thus, alternately, as C is to F, so D (is) to G [Prop. 7.13]. And so, for the same (reasons), as D (is) to G, so E (is) to H . Thus, AT, M, L are contin- uously proportional in the ratio of C to F, and of D to G, and, further, of E to iJ. So let E, H make AT, O, respec- tively, (by) multiplying M. And since A is solid, and C, D, E are its sides, E has thus made A (by) multiplying the (number created) from (multiplying) C, D. And K is the (number created) from (multiplying) C, D. Thus, E has made A (by) multiplying AT. And so, for the same (reasons), H has made B (by) multiplying L. And since £ has made A (by) multiplying K, but has, in fact, also made A^ (by) multiplying M, thus as K is to M, so A (is) to N [Prop. 7.17]. And as K (is) to M, so C (is) to F, and D to G, and, further, E to H. And thus as C (is) to F, and D to G, and F to H, so A (is) to N. Again, since E, H have made N, O, respectively, (by) multiplying M, thus as E is to H, so A^ (is) to O [Prop. 7.18]. But, as E (is) to H, so G (is) to F, and L> to G. And thus as G (is) to F, and D to G, and F to H, so (is) A to N, and AT to O. Again, since H has made O (by) multiplying M, but has, in fact, also made B (by) multiplying L, thus as M (is) to L, so O (is) to B [Prop. 7.17]. But, as M (is) to i, so G (is) to F, and D to G, and F to H. And thus as G (is) to F, and D to G, and F to H, so not only (is) O to F, but also A to AT, and N to O. Thus, A, N, O, B are continuously proportional in the aforementioned ratios of the sides. So I say that A also has to B a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side — that is to say, with respect to (that) the number G (has) to F, or D to G, and, further, F to H. For since A, N, O, B are four continuously proportional numbers, A thus has to B a cubed ratio with respect to (that) A (has) to A^ [Def. 5.10]. But, as A (is) to AT, so it was shown (is) G to F, and D to G, and, further, F to H. And thus A has to B a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side — that is to say, with respect to (that) the number G (has) to F, and D to G, and, further, F to H. (Which is) the very thing it was required to show. Proposition 20 If one number falls between two numbers in mean proportion then the numbers will be similar plane (num- 246 STOIXEIQN rf. ELEMENTS BOOK 8 Auo ydp dpid^wv xwv A, B sic \ieaoz dvdXoyov EjiKiKTETW dpid|ji6c; 6 r- Xeyto, oxi oi A, B 6u.oioi £tutce;8o[ eiaiv dpid^ioi. A i ' A ' ' r ' 1 Z' ' H ' ElX^cpi^waav [yap] eXd)(iaxoi dpidfioi xfiv xov auxov Xoyov eywxcov A, -T 01 ^> ^' todxi? dpa 6 A xov A \LZTpel xdi 6 E xov T. oadxic; 8r) 6 A xov A ^.expeT, xoaauxai ^iovd8ec; eaxwaav ev tw Z' 6 Z dpa xov A TioXXaTiXaaidaac; xov A TieTioir)xev. uoxe 6 A euiTieBoc; eaxiv, TiXeupai 8s auxou oi A, Z. TidXiv, excel oi A, E eXd)(iaxo[ eiai xwv xov auxov Xoyov exovxov xoTc; T, B, iadxic; dpa 6 A xov T [ls- xpeT xal 6 E xov B. oadxic; 5f] 6 E xov B ^lexpeT, xoaaDxai ^.ovdSec; eaxwaav ev iu H. 6 E dpa xov B ^.expel xaxd xdc; ev xQ H [iovd8a<;- 6 H dpa xov E TioXXaTiXaaidaac; xov B TCTCoirjxev. 6 B dpa CTUTieSoc; laxi, TiXeupai 8e auxou eiaiv oi E, H. oi A, B dpa etutisBoi eiaiv dpid^toi. Xeyto 8rj, oti xai o(i.oioi. ETiei yap ° Z xov ^tev A TioXXaTiXaaidaac; xov A TceTioir]xev, xov Be E TioXXaTiXaaidaac; xov T TieTcoirjxev, eaxiv dpa cbc; 6 A Tipoc; xov E, ouxwc; 6 A Tipoc; xov T, xouxeaxiv 6 T Tipoc; xov B. TidXiv, enei 6 E exdxepov xwv Z, H TioXXaTiXaaidaac; xouc; T, B TiCTio[r)xev, eaxiv dpa 6; 6 Z Tipoc; xov H, ouxwc; 6 T Tipoc; xov B. cbc 8s 6 T Tipoc; xov B, ouxoc; 6 A Tipoc; xov E - xal foe, dpa 6 A Tipoc; xov E, ouxoc; 6 Z Tipoc; xov H - xai evaXXac; (be; 6 A Tipoc; xov Z, ouxwc; 6 E Tipoc; xov H. oi A, B dpa o^ioioi etxltxsSoi dpi'd^.oi eiaiv ai yap TiXeupai auxwv dvdXoyov eiaiv oTiep e5ei BeT^ai. xa'. 'Edv 8uo dpid^icov 8uo jieaoi dvdXoyov e^iTUTixoaiv 6ipvQ[Lol, o^ioioi axepeoi eiaiv oi dpid^ioi. Auo ydp dpid|i65v xwv A, B 8uo [ieaoi dvdXoyov e^miTixexwaav dpidfioi oi T, A- Xeyw, oxi oi A, B 6[ioioi axepeoi eiaiv. EiXrjcp'dtoaav ydp eXd)(iaxoi dpi'djioi xfiv xov auxov Xoyov e^ovxeov xolc; A, T, A xpeTc; oi E, Z, H- oi dpa dxpoi auxCSv oi E, H upcoxoi Tipoc; dXXrjXouc; eiaiv. xai CTiel tuv E, H eTc; ^.eaoc; dvdXoyov ejjiTiCTixoxev dpvd^ioc; 6 Z, oi E, H dpa dpidjio! o^xoioi CTUTie8oi eiaiv. eaxoaav ouv xou fiev bers). For let one number C fall between the two numbers A and B in mean proportion. I say that A and B are similar plane numbers. A i D 1 Ci 1 F i 1 G' ' [For] let the least numbers, D and E, having the same ratio as A and C have been taken [Prop. 7.33]. Thus, D measures A as many times as E (measures) C [Prop. 7.20]. So as many times as D measures A, so many units let there be in F. Thus, F has made A (by) multiplying D [Def. 7.15]. Hence, A is plane, and D, F (are) its sides. Again, since D and E are the least of those (numbers) having the same ratio as C and B, D thus measures C as many times as E (measures) B [Prop. 7.20]. So as many times as E measures B, so many units let there be in G. Thus, E measures B ac- cording to the units in G. Thus, G has made B (by) mul- tiplying E [Def. 7.15]. Thus, B is plane, and E, G are its sides. Thus, A and B are (both) plane numbers. So I say that (they are) also similar. For since F has made A (by) multiplying D, and has made C (by) multiplying E, thus as D is to E, so A (is) to C — that is to say, C to B [Prop. 7.17]. t Again, since E has made C, B (by) multi- plying F, G, respectively, thus as F is to G, so C (is) to B [Prop. 7.17]. And as C (is) to B, so D (is) to E. And thus as D (is) to E, so F (is) to G. And, alternately, as D (is) to F, so E (is) to G [Prop. 7.13]. Thus, A and B are similar plane numbers. For their sides are proportional [Def. 7.21]. (Which is) the very thing it was required to show. C. Furthermore, it is not necessary to show that D : E :: A : C, Proposition 21 If two numbers fall between two numbers in mean proportion then the (latter) are similar solid (numbers) . For let the two numbers C and D fall between the two numbers A and B in mean proportion. I say that A and B are similar solid (numbers) . For let the three least numbers E, F, G having the same ratio as A, C, D have been taken [Prop. 8.2]. Thus, the outermost of them, E and G, are prime to one an- other [Prop. 8.3]. And since one number, F, has fallen (between) E and G in mean proportion, E and G are t This part of the proof is defective, since it is not demonstrated that F x E = because this is true by hypothesis. 247 STOIXEIQN rf. ELEMENTS BOOK 8 E rcXeupai oi 9, K, xou 8e H oi A, M. cpavepov apa eaxlv ex xou Tcpo toutou, oti oi E, Z, H e^fjc; eiaiv dvdXoyov ev xe tu xou 9 Ttpoc; tov A Xoycp xod tc5 xou K rcpoi; tov M. xal excel oi E, Z, H eXd)(iaTo[ eiai xwv tov auxov Xoyov e)(6vTWv xou; A, T, A, xai saxiv laov to TiXfj'Oog tGv E, Z, H tw icX^^ei tGv A, T, A, 81' i'aou apa eotiv &>c, 6 E Ttpoc; tov H, ouTtog 6 A rcpoc; tov A. oi Se E, H rcpcoTOi, oi 8e 7ip«Toi xal eXd/iaTOi, oi Se eXd)(ioToi (iCTpouai Toug tov auTov Xoyov exovTac; auTolc; iadxig o ts jjiei^cov tov jiei^ova xai 6 eXdaawv tov eXdaaova, toutsotiv o tc rjyou^ievoc tov r)you[ievov xai 6 erco^evoc; tov ejco^ievov iadxic; apa 6 E tov A [iSTpsi xal 6 H tov A. oadxic; 8f] 6 E tov A [Lexpei, ToaauTai [iovd8e<; eaTwaav ev tw N. 6 N apa tov E TcoXXaicXaaidaac tov A tctco^xsv. 6 8e E eaTiv 6 ex xuv 9, K- 6 N apa tov ex t5v 9, K TtoXXanXaaidaai; tov A iceTcofyxev. axepeoc; apa laxlv 6 A, xcXeupal 8e auToO eiaiv oi 9, K, N. rcdXiv, enel oi E, Z, H eXd/iaToi eiai tGv tov auTov Xoyov e/ovTiov to!<; T, A, B, iadxi? apa 6 E tov T ^.ETpei xai 6 H tov B. oadxu; 8r) 6 E tov T [iCTpel, ToaauTai ^.ovdBec; eaTwaav ev xfi S. 6 H apa tov B ^CTpeT xaTa Tag ev t& S [iovdSac;- 6 S apa tov H TroXXanXaaidaac; tov B Tterconqxev. 6 8s: H eaTiv 6 ex twv A, M' 6 S apa tov ex tGv A, M noXXanXaaidaat; tov B TteTio[r]xev. aTepeoc; apa eaTiv 6 B, rcXeupai Se auTou eiaiv oi A, M, £• oi A, B apa cnxpeoi eiaiv. A i 1 ©i 1 r ' ' K' ' A 1 ' N' ' B ' ' Ei ' A 1 1 Z ' ' Mi ' H 1 1 H 1 1 Aeyw [8r]], oti xai o^toioi. creel yap oi N, H tov E tcoX- XarcXaaidaavTec tou<; A, T Tcercoiiqxaaiv, eaTiv apa cbc; 6 N rcpoc; tov S, 6 A rcpoc; tov T, toutcotiv 6 E rcpoc; tov Z. dXX' cbc; 6 E rcpoc; tov Z, 6 9 rcpoc; tov A xal 6 K rcpoc; tov M- xai (b? apa 6 9 rcpoc; tov A, outwc; 6 K rcpoc; tov M xai 6 N Ttpoc tov H. xai eiaiv oi piev 9, K, N rcXeupai tou A, thus similar plane numbers [Prop. 8.20]. Therefore, let H, K be the sides of E, and L, M (the sides) of G. Thus, it is clear from the (proposition) before this (one) that E, F, G are continuously proportional in the ratio of H to L, and of K to M. And since E, F, G are the least (num- bers) having the same ratio as A, C, D, and the multitude of E, F, G is equal to the multitude of A, C, D, thus, via equality, as E is to G, so A (is) to D [Prop. 7.14]. And E and G (are) prime (to one another), and prime (numbers) are also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (num- bers) measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser — that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures A the same number of times as G (measures) D. So as many times as E measures A, so many units let there be in N. Thus, N has made A (by) multiplying E [Dei. 7.15]. And E is the (number created) from (multiplying) H and K. Thus, has made A (by) multiplying the (number created) from (multiplying) H and K. Thus, A is solid, and its sides are H , K, N. Again, since E, F, G are the least (numbers) having the same ratio as C, D, B, thus E measures C the same number of times as G (measures) B [Prop. 7.20]. So as many times as E measures C, so many units let there be in O. Thus, G measures B according to the units in O. Thus, O has made B (by) multiplying G. And G is the (number created) from (multiplying) L and M. Thus, O has made B (by) multiplying the (number cre- ated) from (multiplying) L and M. Thus, B is solid, and its sides are L, M, O. Thus, A and B are (both) solid. Ah O L h Mh Gh Oh [So] I say that (they are) also similar. For since N, O have made A, C (by) multiplying E, thus as A^ is to O, so A (is) to C — that is to say, E to F [Prop. 7.18]. But, as E (is) to F, so H (is) to L, and K to M. And thus as H (is) to L, so K (is) to M, and N to O. And H, K, N are the sides of A, and L, M, O the sides of B. Thus, A and 248 STOIXEIQN rf. ELEMENTS BOOK 8 oi 8e S, A, M TtXeupai xou B. oi A, B dpa dpi-d^oi o^oioi axepeoi etcnv onep eSsi 8eT<;ai. t The Greek text has "O, L, M", which is obviously a mistake. x(3'. 'Edv xpelc dpi%oi e^fjc; dvdXoyov ffioiv, 6 Ss Tipcoxog xexpdytovot; fj, xai 6 xpixo<; xexpdywvoc; Eaxai. A i 1 B i 1 r ' ' "Eaxwaav xpeu; dpid^ioi £E,ffc dvdXoyov oi A, B, T, 6 8e Tipwxoi; 6 A xExpaycovoc saxw Xsyw, oxi xal 6 xpixoc; 6 T xexpdycovoc; saxiv. 'Etc! yap xwv A, r zic, \itaoc, dvdXoyov saxiv dpid^icx; 6 B, oi A, r dpa o^ioioi eiiLKeSoi eiaiv. xexpdytdvoc 8s: 6 A- xexpdywvo<; dpa xai 6 E ojtep e8ei 8eTc;ai. xy'. 'Eav xeaaapec; dpid^oi e^rjc; dvdXoyov Sow, 6 Se upaixog xupog fj, xai 6 xexapxoc xupoc eaxai. A' ■ B i ' r i 1 A | 1 'Eaxwaav xeaaape<; dpid^ioi e<;fj<; dvdXoyov oi A, B, T, A, 6 8s A x6po<; eaxw Xeyco, oxi xai 6 A x6po<; eaxiv. 'Ercei yap xwv A, A 8uo \±eooi dvdXoyov eiaiv dpidjjioi oi B, T, oi A, A dpa opioioi eiai axepeoi dpi%oi. xupoc; 8s 6 A- x6po<; dpa xai 6 A- onep eBei SeT^ai. x5'. 'Eav 8uo dpiiD^toi Ttpoc; dXXf]Xou<; Xoyov e/waiv, ov xexpdywvoc; dpnSuoc; xpoc; xexpdyovov dpid^tov, 6 8e Ttpwxoc; xexpdyiovoc; fj, xai 6 Seuxepoc; xexpdytovoc; eaxai. A i ' i r i ' B i 1 A i 1 Auo yap dpi%ol oi A, B Ttpoc; dXXiqXouc; Xoyov B are similar solid numbers [Def. 7.21]. (Which is) the very thing it was required to show. Proposition 22 If three numbers are continuously proportional, and the first is square, then the third will also be square. A' 1 B' 1 C' ' Let A, B, C be three continuously proportional num- bers, and let the first A be square. I say that the third C is also square. For since one number, B, is in mean proportion to A and C, A and C are thus similar plane (numbers) [Prop. 8.20]. And A is square. Thus, C is also square [Def. 7.21]. (Which is) the very thing it was required to show. Proposition 23 If four numbers are continuously proportional, and the first is cube, then the fourth will also be cube. A' ' B' ' C' 1 D 1 Let A, B, C, D be four continuously proportional numbers, and let A be cube. I say that D is also cube. For since two numbers, B and C, are in mean propor- tion to A and D, A and D are thus similar solid numbers [Prop. 8.21]. And A (is) cube. Thus, D (is) also cube [Def. 7.21]. (Which is) the very thing it was required to show. Proposition 24 If two numbers have to one another the ratio which a square number (has) to a(nother) square number, and the first is square, then the second will also be square. A i 1 C' ' B i 1 D 1 For let two numbers, A and B, have to one another 249 STOIXEIQN rf. ELEMENTS BOOK 8 e/exGjaav, ov xexpdycovoc; dpid^oc; 6 T 7tp6<; xexpdyovov dpi-Ojiov xov A, 6 8e A xexpdycovoc eaxto - Xeyw, on xal 6 B xexpdycovoc eaxiv. 'Eitei yap oi T, A xexpdycovoi eiaiv, oi T, A dpa ouoioi eraueSoi eiaiv. xcov T, A dpa etc ^jieaoc dvdXoyov e^tTUKxei dpid^oc. xa( eaxiv cbc 6 T Ttpoc xov A, 6 A Ttpoc xov B- xal xov A, B dpa eTc; jieaoc dvdXoyov e^jutmxei dpid^ioc;. xa[ eaxiv 6 A xexpdycovoc;- xal 6 B dpa xexpdycovoc eaxiv oTtep e8ei SeTcai. the ratio which the square number C (has) to the square number D. And let A be square. I say that B is also square. For since C and D are square, C and D are thus sim- ilar plane (numbers). Thus, one number falls (between) C and D in mean proportion [Prop. 8.18]. And as C is to D, (so) A (is) to B. Thus, one number also falls (be- tween) A and B in mean proportion [Prop. 8.8]. And A is square. Thus, B is also square [Prop. 8.22]. (Which is) the very thing it was required to show xs'. 'Edv 8uo dprd^iol npoc dXXf|Xo\Jc Xoyov e/coaiv, ov xuf3o<; dpi^oc rcpoc xu[3ov dpiduov, 6 8e itpcoxoc xu[3oc; rj, xal 6 Beuxepoc xupoc eaxai. a i 1 r ' ' E' ' Z' 1 B i 1 A i 1 Auo yap apidjjioi oi A, B Ttpoc dXXiqXouc Xoyov e^excoaav, ov xupoc dpidjioc 6 T Ttpoc xu^ov dpidjiov xov A, xupoc 8e eaxco 6 A- Xeyco [8r|], oxi xal 6 B xupoc eaxiv. 'End yap oi T, A xu[3oi eiaiv, oi T, A ouoioi axe- peoi eiaiv xcov T, A dpa 8uo ueaoi dvdXoyov euTUTtxouaiv dprduoi. oaoi 8e eic xouc T, A uexacu xaxd xo auvexec dvdXoyov euTUTtxouaiv, xoaouxoi xal eic xouc xov auxov Xoyov e)(ovxac auxolc' coaxe xal xcov A, B 8uo ueaoi dvdXoyov euTUTtxouaiv apiduoi. euTtiTtxexcoaav oi E, Z. eitei ouv xeaaapec dpi-d^ol oi A, E, Z, B ei;rjc dvdXoyov eiaiv, xai eaxi xupoc 6 A, xupoc dpa xal 6 B- oitep eBei 8eTc;ai. Proposition 25 If two numbers have to one another the ratio which a cube number (has) to a(nother) cube number, and the first is cube, then the second will also be cube. A i ' C' 1 Ei ' F ' ' B i 1 D 1 For let two numbers, A and B, have to one another the ratio which the cube number C (has) to the cube number D. And let A be cube. [So] I say that B is also cube. For since C and D are cube (numbers), C and D are (thus) similar solid (numbers). Thus, two numbers fall (between) C and D in mean proportion [Prop. 8.19]. And as many (numbers) as fall in between C and D in continued proportion, so many also (fall) in (between) those (numbers) having the same ratio as them (in con- tinued proportion) [Prop. 8.8]. And hence two numbers fall (between) A and B in mean proportion. Let E and F (so) fall. Therefore, since the four numbers A, E, F, B are continuously proportional, and A is cube, B (is) thus also cube [Prop. 8.23]. (Which is) the very thing it was required to show. xt'. Proposition 26 Oi ouoioi eTUTieSoi dpii9uoi Ttpoc dXXiqXouc Xoyov exou- Similar plane numbers have to one another the ratio aiv, ov xexpdycovoc dpi$u6c Ttpoc xexpdycovov dpnf)uov. which (some) square number (has) to a(nother) square number. A' 1 A 1 A' 1 Di 1 r> 1 E' ' C' 1 B ' 1 Z 1 B 1 F ^ "Eaxcoaav ouoioi ctuttcSoi apvduoi oi A, B- Xeyco, oxi Let A and B be similar plane numbers. I say that A 6 A Ttpoc xov B Xoyov e/ei, ov xexpdycovoc dpiiSuoc Ttpoc has to B the ratio which (some) square number (has) to 250 STOIXEIQN rf. ELEMENTS BOOK 8 xexpdywvov dpiduov. Tkel ydp oi A, B ojioioi eiuTteSoi eiaiv, t«v A, B dpa sic; \isaoc, dvdXoyov cjituttxci dpi%6c;. e^ranxexco xdi eaxto 6 r, xdi eiXr|cpiL>Gjaav eXd)(iaxoi dpid^ioi x65v xov auxov Xoyov e/ovxcov xou; A, r, B oi A, E, Z- oi dpa dxpoi auxcov oi A, Z xexpdycovoi daw. xdi enei eaxiv (be 6 A rcpoc; xov Z, ouxcoc; 6 A Ttpoc; xov B, xai eiaiv oi A, Z xexpdywvoi, 6 A dpa rcpoc; xov B Xoyov exei, ° v te^pdycovoc dpid^io? -rcpoc; xexpdycovov dpid^tov orcep eBei BeT^ai. Oi o^toioi axepeoi dpid^toi 7ip6<; dXXr|Xou<; Xoyov e)(ou- olv, ov xupoc; dpid^toc; 7tp6<; xupov dpid^tov. A 1 ' Ei ' r ' ' z ' ' A 1 1 H' ' B i 1 ©| 1 'Eaxcoaav o^toioi axepeoi dpid^toi oi A, B' Xeyw, oxi 6 A npbc, xov B Xoyov e/ei, ov xufioc, dpid^oc; Kpo<; xupov dpid^iov. Tkei ydp oi A, B o^ioioi axepeoi eiaiv, xwv A, B dpa 8uo ^xeaoi dvdXoyov e^KiKxouaiv dpid^oi. e^XTiiTixexwaav oi T, A, xai eiXf](pTf)waav eXd)(iaxoi dpid^toi xwv xov auxov Xoyov exovxwv xolc A, T, A, B laoi auxoTg xo nXfj'doc; oi E, Z, H, 9- oi dpa dxpoi auxwv oi E, xOpoi eiaiv. xai eaxiv cbc 6 E 7tp6<; xov 6, ouxw<; 6 A 7tp6<; xov B- xai 6 A dpa npbc, xov B Xoyov exei, Sv x6po<; dpi'djioc; npoc, xupov dpi%6v oTiep e8ei SeT^ai. a(nother) square number. For since A and B are similar plane numbers, one number thus falls (between) A and B in mean propor- tion [Prop. 8.18]. Let it (so) fall, and let it be C. And let the least numbers, D, E, F, having the same ratio as A, C, B have been taken [Prop. 8.2]. The outermost of them, D and F, are thus square [Prop. 8.2 corr.]. And since as D is to F, so A (is) to B, and D and F are square, A thus has to B the ratio which (some) square number (has) to a(nother) square number. (Which is) the very thing it was required to show. Proposition 27 Similar solid numbers have to one another the ratio which (some) cube number (has) to a(nother) cube num- ber. A i 1 E i 1 Ci 1 F i 1 Di 1 G' 1 B 1 Hi 1 Let A and B be similar solid numbers. I say that A has to B the ratio which (some) cube number (has) to a(nother) cube number. For since A and B are similar solid (numbers), two numbers thus fall (between) A and B in mean proportion [Prop. 8.19]. Let C and D have (so) fallen. And let the least numbers, E, F, G, H, having the same ratio as A, C, D, B, (and) equal in multitude to them, have been taken [Prop. 8.2]. Thus, the outermost of them, E and H, are cube [Prop. 8.2 corr.]. And as E is to H, so A (is) to B. And thus A has to B the ratio which (some) cube number (has) to a(nother) cube number. (Which is) the very thing it was required to show. 251 252 ELEMENTS BOOK 9 Applications of Number 77ieory| tThe propositions contained in Books 7-9 are generally attributed to the school of Pythagoras. 253 ETOIXEIfiN fl'. ELEMENTS BOOK 9 a . 'Edv 860 6[ioioi eiuTieSoi dpidjiol TtoXXanXaaidaavxec; dXXr]Xouc; noiwai xiva, 6 yevo^ievoc; xexpdycovoc; eaxai. A' 1 B ' ' r 1 1 A' ' "Eaxwaav 860 8(ioioi eiuneSoi dpi%ol 01 A, B, xal 6 A xov B TtoXXaTtXaaidaac; xov T Ttoieixw Xeyw, oxi 6 T xexpdycovo? eaxiv. c yap A eauxov rcoXXaTtXaaidaa? xov A uoieixw. 6 A dpa xexpdytovo? eaxiv. inel ouv 6 A eauxov [lev TtoXXa- TtXaaidaac; xov A Tte7toir)xev, xov 8s B rcoXXaTtXaaidaac; xov r TiSKOi/jxev, eaxiv dpa 6 A 7tp6<; xov B, oux«<; 6 A 7ipo<; xov T. xal etiei 01 A, B o^ioioi CTUTteSoi eiaiv dpid^ol, x£Sv A, B dpa elc, [leooc, dvdXoyov IjiitiinEi dpid^to?. eav 8e 860 dpid^iGv [icxa^u xaxa xo auvexe? dvdXoyov e^miKxoaiv dpi'd^oi, oaoi ei? auxou? e^TUTtxouai, xoaouxoi xal el? xou? xov auxov Xoyov exovxag- &axe xal xwv A, T eT? [leaoq, dvdXoyov e^tiitixei dpid^o?. xal eaxi xexpdywvo? 6 A- xexpdyovo? dpa xal 6 E oitep e§ei 8eTc;ai. P'- 'Eav 860 dpid^iol 7toXXaTtXaaidaavxe<; dXXr]Xou<; itoicoai xsxpdywvov, 0^.0101 ekikeBoi eiaiv dpid^.01. A' 1 B 1 1 r 1 1 A' ■ TEaxwaav 860 dpi-duol 01 A, B, xal 6 A xov B KoXXa- TtXaaidaac; xexpdycovov xov T noislxw Xeyw, oxi oi A, B ojioioi eraTieSoi eiaiv dpidjioi. yap A eauxov TroXXanXaaidaac; xov A Tioieixor 6 A dpa xexpdytovo? eaxiv. xal end 6 A eauxov (lev TtoXXa- TtXaaidaac; xov A Kenoi/jxev, xov 8e B TioXXanXaoidaat; xov r nenonqxev, eaxiv dpa 6 A Ttpoc; xov B, 6 A Ttpoc; xov T. xal stcI 6 A xexpdywvog eaxiv, dXXd xal 6 T, oi A, T dpa ofioioi eniTieSoi eiaiv. xfiv A, T dpa si? [isaog dvdXoyov Proposition 1 If two similar plane numbers make some (number by) multiplying one another then the created (number) will be square. A' ' B 1 C' ' D 1 Let A and B be two similar plane numbers, and let A make C (by) multiplying B. I say that C is square. For let A make D (by) multiplying itself. D is thus square. Therefore, since A has made D (by) multiply- ing itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since A and B are similar plane numbers, one number thus falls (be- tween) A and B in mean proportion [Prop. 8.18]. And if (some) numbers fall between two numbers in continued proportion then, as many (numbers) as fall in (between) them (in continued proportion), so many also (fall) in (between numbers) having the same ratio (as them in continued proportion) [Prop. 8.8]. And hence one num- ber falls (between) D and C in mean proportion. And D is square. Thus, C (is) also square [Prop. 8.22]. (Which is) the very thing it was required to show. Proposition 2 If two numbers make a square (number by) multiply- ing one another then they are similar plane numbers. A' ' B 1 C' ' D 1 Let A and B be two numbers, and let A make the square (number) C (by) multiplying B. I say that A and B are similar plane numbers. For let A make D (by) multiplying itself. Thus, D is square. And since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since D is square, and C (is) also, D and C are thus similar plane numbers. Thus, one (number) falls (between) D and C in mean propor- 254 ETOIXEIfiN fl'. ELEMENTS BOOK 9 ejitutctei. xa[ eaxiv cbc; 6 A Ttpoc; xov T, ouxcoc; 6 A Ttpoc; xov B- xod xfiv A, B apa elc; \±£ooz dvdXoyov e^miuxei. edv 8s 8uo dpi%cov sic, [Leaoc, dvdXoyov e^tutitt], o^toioi stutieSoi eiaiv [oi] dpid^ioi- oi apa A, B o^toioi eiaiv eiuTieBoi- oitep eSei 8eTc;ai. Y'- 'Edv xupoc; dpn^oc; eauxov TtoXXarcXaaidaac; uoifj xiva, 6 yevojievoc; xupoc; eaxai. A i 1 B i 1 r i 1 A i ' Kupoc; yap dpidjjidc; 6 A eauxov KoXXaTtXaaidaac; xov B itoieixco- Xeyo, oxi 6 B xupoc; eaxiv. ElXf](p , dcL) yap xou A TtXeupd 6 T, xal 6 T eauxov KoXXa- TtXaaidaac; xov A noieixco. cpavepov 8Vj eaxiv, oxi 6 T xov A TioXXaiiXaaidaac; xov A TteTCofyxev. xal cuei 6 T eauxov tioX- XaTtXaaidaac; xov A Ti£7io[r)X£v, 6 T apa xov A ^.expel xaxa xdc; ev auxo \xov6&u.c,. dXXd ^irjv xal f\ [iovdc xov T ^.expeT xaxa xdc; ev auxfi uovdBac;- eaxiv apa cbc; f] \xovac, Tipoc; xov r, 6 r Ttpoc; xov A. TtdXiv, CTiel 6 T xov A TioXXaTtXaaidaac; xov A 7i£iio[r]xev, 6 A apa xov A jiexpeT xaxa xdc ev xw T ^iovd8ac;. ^texpel 8e xal f) ^tova<; xov T xaxa xdc; ev auxfi ^tovd8ac;' eaxiv apa 6c; f] ^lovdc; Tipoc; xov T, 6 A Tipoc; xov A. dXX' wc; f) [iovdc; Tipoc; xov T, 6 T Tipoc; xov A- xal dbc; apa f) jiovdc; Tipoc; xov T, ouxcoc; 6 T Tipoc; xov A xal 6 A Tipoc; xov A. xrjc; apa jiovdSoc; xal xou A dpid^iou 8uo ^.eaoi dvdXoyov xaxa xo auvexec; ejiTCCTixoxaaiv dpiiJ^ol oi T, A. TtdXiv, etc el 6 A eauxov TioXXaTtXaaidaac; xov B TieTtoirjxev, 6 A apa xov B (jiexpeT xaxa xdc; ev auxw ^ovd8ac;' [icxpel 8e xal f) jiovdc; xov A xaxa xdc; ev auxCS jiovdBac;- eaxiv apa (be; f] ^.ovdc; Tipoc; xov A, 6 A Tipoc; xov B. xfjc; 8e [lovdBoc; xal xou A 8uo ^.eaoi dvdXoyov ejiiceuxoxaaiv dpi'd^oi' xal x£>v A, B apa Buo \ieaoi dvdXoyov ejiiceaouvxai dpi'd^ioi. edv 8e 8uo dpi-d^wv 8uo ^teaoi dvdXoyov e^miiixwaiv, 6 8e TipGxoc; xupoc; fj, xal 6 Beuxepoc; xupoc; eaxai. xa( eaxiv 6 A xupoc xal 6 B apa xupoc; eaxiv oTtep eSei Belial. 5'. 'Edv xupoc; dpid^ioc; xupov dpid^iov TioXXaTtXaaidaac; Ttoifj xiva, 6 yev6[ievoc; xupoc; eaxai. tion [Prop. 8.18]. And as D is to C, so A (is) to B. Thus, one (number) also falls (between) A and B in mean pro- portion [Prop. 8.8]. And if one (number) falls (between) two numbers in mean proportion then [the] numbers are similar plane (numbers) [Prop. 8.20]. Thus, A and B are similar plane (numbers). (Which is) the very thing it was required to show. Proposition 3 If a cube number makes some (number by) multiply- ing itself then the created (number) will be cube. A' 1 Bi 1 Ci ' D 1 For let the cube number A make B (by) multiplying itself. I say that B is cube. For let the side C of A have been taken. And let C make D by multiplying itself. So it is clear that C has made A (by) multiplying D. And since C has made D (by) multiplying itself, C thus measures D according to the units in it [Def. 7.15]. But, in fact, a unit also mea- sures C according to the units in it [Def. 7.20] . Thus, as a unit is to C, so C (is) to D. Again, since C has made A (by) multiplying D, D thus measures A according to the units in C. And a unit also measures C according to the units in it. Thus, as a unit is to C, so D (is) to A. But, as a unit (is) to C, so C (is) to D. And thus as a unit (is) to C, so C (is) to D, and D to A. Thus, two numbers, C and D, have fallen (between) a unit and the number A in continued mean proportion. Again, since A has made B (by) multiplying itself, A thus measures B according to the units in it. And a unit also measures A according to the units in it. Thus, as a unit is to A, so A (is) to B. And two numbers have fallen (between) a unit and A in mean proportion. Thus two numbers will also fall (be- tween) A and B in mean proportion [Prop. 8.8]. And if two (numbers) fall (between) two numbers in mean pro- portion, and the first (number) is cube, then the second will also be cube [Prop. 8.23]. And A is cube. Thus, B is also cube. (Which is) the very thing it was required to show. Proposition 4 If a cube number makes some (number by) multiply- ing a(nother) cube number then the created (number) 255 ETOIXEIfiN fl'. ELEMENTS BOOK 9 A' ' B i 1 r ' ' a i 1 Kupoc; yap dpi'd^oc; 6 A xupov dpid^iov xov B TtoXXa- TiXaoidaac; xov T tcoieitw Xeyto, oxi 6 T xupoc; eaxiv. c O yap A eauxov TtoXXarcXaaidaac; xov A Ttoieixw 6 A dpa xupoc; eaxiv. xal ercel 6 A eauxov uev TtoXXa- TtXaaidaac; xov A TteTtoirjxev, xov 8s B rcoXXaTiXaaidaac; xov r K£7io[r)X£v, eaxiv dpa ox; 6 A rcpoc; xov B, ouxoc; 6 A Ttpoc; xov T. xal ETtsi oi A, B xupoi eiaiv, o^ioioi axepeoi eiaiv oi A, B. x«v A, B dpa 80o [leaoi dvdXoyov ejiTunxouaiv dpi%o£- Saxe xal x«v A, r Buo [isaoi dvdXoyov ejjmeaouvxai dpii9|jioL. xa[ eaxi xupoc; 6 A- xupoc; dpa xal 6 T- onep eSei 8e"i£ai. e'. °Eav xupog dpi-d^oc; dpi%6v xiva TioXXaTtXaaidaac; xupov Tioifj, xal 6 TtoXXaTiXaaiatydeic; xupoc; eaxai. A i 1 B i 1 r i 1 A' ■ Kupoc; yap dpi'djioc; 6 A apiduov xiva xov B TtoXXa- TtXaaidaac; xupov xov T noieixw Xeyw, oxi 6 B xupoc; eaxiv. c yap A eauxov TtoXXaitXaaidaac; xov A Ttoieixco - xupoc; dpa saxiv 6 A. xal ercel 6 A eauxov ^tev TCoXXanXaaidaac; xov A 7ieTio(r]xev, xov 8e B TtoXXaitXaaidaac; xov T TiCTtoirjxev, eaxiv dpa (be; 6 A Ttpoc; xov B, 6 A Ttpoc; xov 1\ xal excel oi A, T xupoi eioiv, ojioioi axepeoi eioiv. xfiv A, T dpa Suo [ieaoi dvdXoyov eu-iciicxouaiv dpid^oi. xai eaxiv wc; 6 A jepoe; xov r, ouxwc; 6 A Tipoc; xov B- xai xwv A, B dpa 8uo [icooi dvdXoyov ejiTUTixouaiv dpidjioi. xai eaxi xupoc; 6 A- xupoc; dpa eaxi xal 6 B- onep e8ei 8el^ai. 'Eav dpidfjioc; eauxov TcoXXaTiXaaidaac; xupov noifj, xal will be cube. A' ' B' 1 Ci 1 Di 1 For let the cube number A make C (by) multiplying the cube number B. I say that C is cube. For let A make D (by) multiplying itself. Thus, D is cube [Prop. 9.3]. And since A has made D (by) multi- plying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since A and B are cube, A and B are similar solid (numbers). Thus, two numbers fall (between) A and B in mean pro- portion [Prop. 8.19]. Hence, two numbers will also fall (between) D and C in mean proportion [Prop. 8.8]. And D is cube. Thus, C (is) also cube [Prop. 8.23]. (Which is) the very thing it was required to show. Proposition 5 If a cube number makes a(nother) cube number (by) multiplying some (number) then the (number) multi- plied will also be cube. A' 1 B' 1 O ' D' 1 For let the cube number A make the cube (number) C (by) multiplying some number B. I say that B is cube. For let A make D (by) multiplying itself. D is thus cube [Prop. 9.3]. And since A has made D (by) multiply- ing itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17]. And since D and C are (both) cube, they are similar solid (numbers) . Thus, two numbers fall (between) D and C in mean proportion [Prop. 8.19]. And as D is to C, so A (is) to B. Thus, two numbers also fall (between) A and B in mean pro- portion [Prop. 8.8]. And A is cube. Thus, B is also cube [Prop. 8.23]. (Which is) the very thing it was required to show. Proposition 6 If a number makes a cube (number by) multiplying 256 ETOIXEIfiN fl'. ELEMENTS BOOK 9 auxog xu[3o<; eoxai. A' 1 B i 1 r i 1 'Api'd(i6<; yap 6 A eauTov TtoXXaTtXaaidaac; xupov tov B tioieitw Xeyw, oxi xal 6 A xupoc; eaTiv. c yap A tov B TioXXaTiXaaidaag tov T Ttoierao. enel ouv 6 A eauTov ^iev 7ioXXaTtXaaidaa<; tov B TtCTtoirjxev, tov Be: B TioXXaTtXaaidaac; tov V Tienoirjxsv, 6 T dpa xupoc; eaTiv. xal ettsI 6 A eauTov TtoXXaTtXaaidaac; tov B kstioi/jxev, 6 A dpa tov B ^lETpel xaTa Tag ev auTfi jiovdSac;. (iCTpel Be xal f) ^iovd<; tov A xaTa Tag sv auTfi (jiovaBag. eaTiv dpa cbc f] jiovac; 7ip6<; tov A, outw<; 6 A 7tp6<; tov B. xal ensl 6 A tov B TtoXXaTtXaaidaac; tov T TteTTOirjxsv, 6 B dpa tov r (jLexpeT xaTa Tag ev to A ^ovd8a<;. ^.eTpel 8e xal rj ^ovdc tov A xaTa Ta<; ev auTW ^iovd8a<;. eaTiv dpa w<; f] [Lovaz 7ipo<; tov A, outw<; 6 B Ttpoc; tov T. dXX' d><; f] [lovolq npbz tov A, outoc 6 A rcpoc; tov B- xal (i><; dpa 6 A upoc tov B, 6 B npoz tov T. xal eitel ol B, T xupoi eiaiv, ojioioi cnxpeoi eiaiv. twv B, T dpa 8uo [isaoi dvdXoyov eiaiv dpi'd^oi. xai eaTiv 6 B Ttp6<; tov T, 6 A Ttpoc; tov B. xal twv A, B dpa 8uo [Leaoi dvdXoyov eiaiv dpid^ioi. xa[ eaTiv xupoc; 6 B- xupoc; dpa eotI xal 6 A- ouep e8ei Sel^ai. c 'Eav auvdeToc; dpid^toc; dpidjiov Tiva TioXXajcXaaidaac; icoifj Tiva, 6 yevojievoc; oxepebq eaxca. A' 1 B i 1 r i 1 A i ' Ei 1 SuvdsTog yap dpid^ioc; 6 A dpid^ov Tiva tov B TtoXXa- jcXaaidaac; tov T uoieiTW Xeyw, oti 6 T aTepeoc; eaTiv. 'End yap 6 A auvdeToc; eaTiv, Otto dpi^ou tivoc; (ie- Tp/)i9r]aeTai. |ieTpe[ad« utto tou A, xal oadxic; 6 A tov A (iCTpel, ToaaOTai (iovd8e<; eaTwaav sv t£) E. insi ouv 6 A tov A (iETpel xaTa Tag ev tG E |iovd8a<;, 6 E dpa tov A TtoXXaTtXaaidaac; tov A TteitoiTjxev. xal eitel 6 A tov B ttoX- XanXaaidaac; tov T Tieno[r)xev, 6 Se A eaTiv 6 ex to>v A, E, 6 dpa ex t«v A, E tov B TtoXXaitXaaidaac; tov T nenoi/jxev. 6 T dpa aTepeoc; eaTiv, icXeupal 5e auToO eiaiv oi A, E, B- itself then it itself will also be cube. A' 1 B 1 C' 1 For let the number A make the cube (number) B (by) multiplying itself. I say that A is also cube. For let A make C (by) multiplying B. Therefore, since A has made B (by) multiplying itself, and has made C (by) multiplying B, C is thus cube. And since A has made B (by) multiplying itself, A thus measures B according to the units in (A). And a unit also measures A according to the units in it. Thus, as a unit is to A, so A (is) to B. And since A has made C (by) multiplying B, B thus measures C according to the units in A. And a unit also measures A according to the units in it. Thus, as a unit is to A, so B (is) to C. But, as a unit (is) to A, so A (is) to B. And thus as A (is) to B, (so) B (is) to C. And since B and C are cube, they are similar solid (numbers) . Thus, there exist two numbers in mean proportion (between) B and C [Prop. 8.19]. And as B is to C, (so) A (is) to B. Thus, there also exist two numbers in mean proportion (between) A and B [Prop. 8.8]. And B is cube. Thus, A is also cube [Prop. 8.23]. (Which is) the very thing it was required to show. Proposition 7 If a composite number makes some (number by) mul- tiplying some (other) number then the created (number) will be solid. A' ' B 1 C' 1 Di 1 Ei ' For let the composite number A make C (by) multi- plying some number B. I say that C is solid. For since A is a composite (number), it will be mea- sured by some number. Let it be measured by D. And, as many times as D measures A, so many units let there be in E. Therefore, since D measures A according to the units in E, E has thus made A (by) multiplying D [Def. 7.15]. And since A has made C (by) multiplying B, and A is the (number created) from (multiplying) D, E, the (number created) from (multiplying) D, E has thus 257 ETOIXEIfiN fl'. ELEMENTS BOOK 9 onep e8ei SeT^ai. 'Edv and jiovdBoc; ottoooiouv dpi'd^iol e^fjc; dvdXoyov CSaiv, 6 jiev xpixoc dfto xfjt; ^lovdBoc; xexpdycovoc; eaxai xal oi eva SiaXeiTtovxec;, 6 8e xexapxoc; xupoc; xal oi 86o SiaXeinovxec; Ttdvxec;, 6 8s epSojioc; xupoc; djia xal xexpdycovoc; xal oi rcevxe SiaXeiTtovxec;. A' 1 B i 1 r i 1 A | 1 Ei 1 Z' 1 "Eaxwaav duo ^tovdSoc; otioooiouv dpid^ioi e^fjc; dvdXoy- ov oi A, B, r, A, E, Z- Xeyto, oxi 6 \±ev xpixoc; duo xrjc ^ovdSoc; 6 B xexpdytovoc; eaxi xal oi eva SiaXeiTtovxec; Ttdvxec;, 6 8s xexapxoc; 6 T xupoc; xal oi 66o SiaXeiTtovxec; Ttdvxec;, 6 8s epBojjioc; 6 Z xupoc; d^ia xal xexpdycovoc; xal oi Ttevxe SiaXeiTtovxec; Ttdvxec;. 'EtcI yap ectxiv toe; f] jiovdc; Ttpoc; xov A, ouxmc; 6 A Ttpoc; xov B, iadxic; dpa f) ^lovdc; xov A dpidjiov [icxpel xal 6 A xov B. f] Se jiovdc; xov A dpnSjiov jiexpeT xaxa xdc; ev auxfi [iovdBac;- xal 6 A dpa xov B [iexpel xaxa xdc; ev xfi A [lovdSac;. 6 A dpa eauxov TtoXXaTtXaaidaac; xov B 7ienoiT)X£V xexpdycovoc; dpa eaxlv 6 B. xal eTtel oi B, T, A e<;rjc; dvdXoyov eiaiv, 6 8s B xexpdycovoc; eaxiv, xal 6 A dpa xexpdycovoc; eaxiv. Sid xd auxd 8r) xal 6 Z xexpdycovoc; eaxiv. ojioicoc; 8/) 8ei^o[i£v, oxi xal oi eva SiaXeiTtovxec; Ttdvxec; xexpdycovoi eiaiv. Xeyco 6r], oxi xal 6 xexapxoc; and xrjg ^xovdSoc; 6 T xupoc eaxi xal oi 66o SiaXeiTtovxec; Ttdvxec;. inzi yap eoxiv cbc; f] [iomolq Ttpoc; xov A, ouxcoc; 6 B Ttpoc; xov r, iadxic; dpa f\ jiovac; xov A dpid^tov (icxpel xal 6 B xov T. f) Se fiovac; xov A dpid^iov ^.expel xaxa xdc; ev xcp A jiovdSac;- xal 6 B dpa xov T ^texpel xaxa xdc; ev iu A ^tovdSac;- 6 A dpa xov B TtoXXaTtXaaidaac; xov T TteTtonqxev. eTtel ouv 6 A eauxov [ikv TtoXXaTtXaaidaac; xov B TteTtoirjxev, xov Se B TtoXXaTtXaaidaac; xov T TteTtoirjxev, xupoc; dpa eoxiv 6 T. xal eTtel oi T, A, E, Z sZ,f\c, dvdXoyov eiaiv, 6 Se T xupoc; eaxiv, made C (by) multiplying B. Thus, C is solid, and its sides are D, E, B. (Which is) the very thing it was required to show. Proposition 8 If any multitude whatsoever of numbers is continu- ously proportional, (starting) from a unit, then the third from the unit will be square, and (all) those (numbers after that) which leave an interval of one (number), and the fourth (will be) cube, and all those (numbers after that) which leave an interval of two (numbers), and the seventh (will be) both cube and square, and (all) those (numbers after that) which leave an interval of five (num- bers). A' ' B 1 C' ' D 1 Ei ' F i ' Let any multitude whatsoever of numbers, A, B, C, D, E, F, be continuously proportional, (starting) from a unit. I say that the third from the unit, B, is square, and all those (numbers after that) which leave an inter- val of one (number). And the fourth (from the unit), C, (is) cube, and all those (numbers after that) which leave an interval of two (numbers) . And the seventh (from the unit), F, (is) both cube and square, and all those (num- bers after that) which leave an interval of five (numbers) . For since as the unit is to A, so A (is) to B, the unit thus measures the number A the same number of times as A (measures) B [Def. 7.20]. And the unit measures the number A according to the units in it. Thus, A also measures B according to the units in A. A has thus made B (by) multiplying itself [Def. 7.15]. Thus, B is square. And since B, C, D are continuously proportional, and B is square, D is thus also square [Prop. 8.22]. So, for the same (reasons), F is also square. So, similarly, we can also show that all those (numbers after that) which leave an interval of one (number) are square. So I also say that the fourth (number) from the unit, C, is cube, and all those (numbers after that) which leave an interval of two (numbers). For since as the unit is to A, so B (is) to C, the unit thus measures the number A the same number of times that B (measures) C. And the unit measures the 258 ETOIXEIfiN fl'. ELEMENTS BOOK 9 xal 6 Z dpa xupo<; eaxiv. £8e[)cdT] Be: xal xexpdycovoc;' 6 dpa ipBo^iot; duo xfjc [iovd8o<; xupoc; xe eaxi xal xexpdycovo<;. o^lolwc; 5f) Bei^o|jiev, oxi xal oi Ttevxe 8iaXemovxe<; 7tdvxe<; xupoi xe eiai xal xexpdywvor orcep eBei BeT^ai. 0'. 'Eav aTio jiovdBoc; oitoaoiouv e^rjc; xaxa xo auvexec dpi%ol dvdXoyov waiv, 6 Be ^exd xiqv ^xovdBa xexpdycovoc fj, xal oi Xomol ndvxec; xexpdycovoi eaovxai. xal eav 6 (icxa xr]v (jiovdSa xupo<; fj, xal oi XoitcoI Ttdvxec; xupoi eaovxai. A' 1 B i 1 r i 1 A | ' E' 1 Z' ' "Eaxwaav cmo ^iovdBo<; 'tEjff, dvdXoyov 6aoiBr)Ttoxouv dipi'd^ol oi A, B, T, A, E, Z, 6 Be ^texa xf)v ^xovdBa 6 A xexpdywvoc; eaxw Xey«, oxi xal oi Xomol udvxsc xexpdytovoi eaovxai. "Oxi fiev ouv 6 xpixoc duo xfj? jiovdBoc 6 B xexpdy«vo<; eaxi xal oi eva SiaTiXemovxet; Ttdvxec;, SeBeixxar Xeyw [8yj], oxi xal oi Xomol navxec; xexpdycovoi eiaiv. Emel yap oi A, B, T e^fj<; dvdXoyov eiaiv, xai eaxiv 6 A xexpdytovo<;, xal 6 T [apa] xexpdywvoc; eaxiv. TtdXiv, tnei [xal] oi B, T, A e^rj<; dvdXoyov eiaiv, xai eaxiv 6 B xexpdywvo^, xai 6 A [apa] xexpdywvo<; eaxiv. 6^ioi«<; Br) Bei^ouev, oxi xal oi Xoircoi iravxec xexpdyovoi eiaiv. AXXa 8r] eaxco 6 A xupoc Xeyco, oxi xal oi Xomol Ttdvxec xupoi eiaiv. "Oxi [iev ouv 6 xexapxoc; duo xfjc; ^ovdBoc; 6 T xupo<; eaxi xal oi 86o SiaXemovxec 7tdvxe<;, BeBeixxar Xeyco [Br]], oxi xai oi Xomol 7tdvxe<; xupoi eiaiv. end yap eaxiv &>z f) [iova<; npbz xov A, ouxw<; 6 A Ttpoc; xov B, iadxic; apa r\ (iovac xov A [iexpel xai 6 A xov B. f) Be ^ovdc; xov A (iexpeT xaxa xac; ev number A according to the units in A. And thus B mea- sures C according to the units in A. A has thus made C (by) multiplying B. Therefore, since A has made B (by) multiplying itself, and has made C (by) multiplying B, C is thus cube. And since C, D, E, F are continuously pro- portional, and C is cube, F is thus also cube [Prop. 8.23]. And it was also shown (to be) square. Thus, the seventh (number) from the unit is (both) cube and square. So, similarly, we can show that all those (numbers after that) which leave an interval of five (numbers) are (both) cube and square. (Which is) the very thing it was required to show. Proposition 9 If any multitude whatsoever of numbers is continu- ously proportional, (starting) from a unit, and the (num- ber) after the unit is square, then all the remaining (num- bers) will also be square. And if the (number) after the unit is cube, then all the remaining (numbers) will also be cube. A' ' B 1 C' ' D 1 Ei ' F i ' Let any multitude whatsoever of numbers, A, B, C, D, E, F, be continuously proportional, (starting) from a unit. And let the (number) after the unit, A, be square. I say that all the remaining (numbers) will also be square. In fact, it has (already) been shown that the third (number) from the unit, B, is square, and all those (num- bers after that) which leave an interval of one (number) [Prop. 9.8]. [So] I say that all the remaining (num- bers) are also square. For since A, B, C are continu- ously proportional, and A (is) square, C is [thus] also square [Prop. 8.22]. Again, since B, C, D are [also] con- tinuously proportional, and B is square, D is [thus] also square [Prop. 8.22]. So, similarly, we can show that all the remaining (numbers) are also square. And so let A be cube. I say that all the remaining (numbers) are also cube. In fact, it has (already) been shown that the fourth (number) from the unit, C, is cube, and all those (num- bers after that) which leave an interval of two (numbers) 259 ETOIXEIfiN fl'. ELEMENTS BOOK 9 auxo) jiovdBac;' xal 6 A dpa xov B [iexpel xaxd xag ev aux£> (iovdSac 6 A apa eauxov TroXXanXaaidaac; tov B tcseoi/jxev. xat eaxiv 6 A x6po<;. edv 8e x6po<; dpiduoc; eauxov TtoXXa- TtXaaidaac; Ttoirj xiva, 6 yevo^ievo<; x6po<; eaxiv xal 6 B dpa xOpoc eaxiv. xal six si xeaaapec; dpid^tol oi A, B, T, A iZ,f\c, dvdXoyov eiaiv, xa[ eaxiv 6 A xupog, xal 6 A apa xupoc eaxiv. Bid xd auxd Sr] xal 6 E xupoc eaxiv, xal 6^toi«<; oi XoitcoI Kdvxsc xupoi eiaiv oTtep eBei BeT^ai. [Prop. 9.8]. [So] I say that all the remaining (numbers) are also cube. For since as the unit is to A, so A (is) to B, the unit thus measures A the same number of times as A (measures) B. And the unit measures A according to the units in it. Thus, A also measures B according to the units in (A) . A has thus made B (by) multiplying it- self. And A is cube. And if a cube number makes some (number by) multiplying itself then the created (number) is cube [Prop. 9.3]. Thus, B is also cube. And since the four numbers A, B, C, D are continuously proportional, and A is cube, D is thus also cube [Prop. 8.23]. So, for the same (reasons), E is also cube, and, similarly, all the remaining (numbers) are cube. (Which is) the very thing it was required to show. i . 'Edv and ^ovd§o<; otcoctoioOv dpiif)[iol [ec;fj<;] dvdXoyov Gaiv, 6 Be ^.exd x/]v jiovdBa \±r\ fj xexpdytovoc;, ou5' aXXoc; ouSelc; xexpdycovoc eaxai x w P^ TO ° tpttou duo xrjg jiovdSoc; xal xwv eva BiaXemovxwv jcdvxwv. xal edv 6 jiexd x/]v uovdSa xupoc [ir] fj, o05e aXXo<; ouBelc; xOpoc eaxai x w P^ xoO xexdpxou dno xfjc; uovdSoc xal x«v 8uo SiaXeinovxov ndvxwv. A' 1 B i 1 r i 1 A | ' E' 1 Z' 1 TCaxcoaav duo ^tovdSoc; e^fj<; dvdXoyov 6aoi8r)n;oxouv dpid^tol oi A, B, T, A, E, Z, 6 uexd x/]v ^tovdSa 6 A [iy] eaxw xexpdywvoc Xey«, oxi ouSe akXoc, o05elc; xexpdycovoc; eaxai x w P^ TO ° tpfrou dico xr]<; uovdBoc; [xal xwv eva 8ia- Xeikovxwv] . EE yap 5uvax6v, eaxw 6 T xexpdywvoc;. eaxi 8e xal 6 B xexpdywvoc oi B, T apa 7ipo<; dXXr|Xou<; Xoyov exou- aiv, ov xexpdytovoc dpidjjioc; Ttpoc xexpdywvov dpid^iov. xai eaxiv &>z 6 B Ttpoc; xov T, 6 A Ttpoc; xov B' oi A, B dpa Ttpoc; dXXrjXouc; Xoyov exouaiv, ° v xexpdytovoc; dpid^ioc; Ttpoc; xexpdyiovov dpi%6v uoie oi A, B ouoioi CTtiTteSoi eiaiv. xai eaxi xexpdytovoc; 6 B- xexpdycovoc; apa eaxi xal 6 A- OTtep oux unexeixo. oux apa 6 T xexpdytovoc; eaxiv. o^ioitoc; 8f) Bei^o^tev, oxi oOS' dXXoc; ouSelc; xexpdyovoc eaxi x w P^ Proposition 10 If any multitude whatsoever of numbers is [continu- ously] proportional, (starting) from a unit, and the (num- ber) after the unit is not square, then no other (number) will be square either, apart from the third from the unit, and all those (numbers after that) which leave an inter- val of one (number) . And if the (number) after the unit is not cube, then no other (number) will be cube either, apart from the fourth from the unit, and all those (num- bers after that) which leave an interval of two (numbers). A' ' B 1 C' ' D 1 Ei 1 F i ' Let any multitude whatsoever of numbers, A, B, C, D, E, F, be continuously proportional, (starting) from a unit. And let the (number) after the unit, A, not be square. I say that no other (number) will be square ei- ther, apart from the third from the unit [and (all) those (numbers after that) which leave an interval of one (num- ber)]. For, if possible, let C be square. And B is also square [Prop. 9.8]. Thus, B and C have to one another (the) ratio which (some) square number (has) to (some other) square number. And as B is to C, (so) A (is) to B. Thus, A and B have to one another (the) ratio which (some) square number has to (some other) square number. Hence, A and B are similar plane (numbers) 260 ETOIXEIfiN fl'. ELEMENTS BOOK 9 xou xplxou duo xfjc; ^xovdBoc; xal xfiv eva BiaXeiTxovxojv. AXXd Bf] (jltq eaxw 6 A xupoc;. Xeyw, oxi o08' dXXoc; ouBelc; xupoc; eaxai x w P^ TO ° texdpxou duo xfjc; jiovdBoc; xal xov Buo BiaXemovxwv. El yap Buvaxov, saxw 6 A xupoc;. eaxi Be: xal 6 T xupoc xexapxoc; yap feaxiv duo xfjc; [LovaSoq,. xal eaxiv cbc; 6 T Tipoc; xov A, 6 B npbc, xov E xal 6 B apa Tipoc; xov T Xoyov exei, 6v xOpoi; Tipoc; xupov. xal eaxiv 6 T xupoc xal 6 B apa xOpoc eaxlv. xal inzi saxiv dbc; f] ^iovd<; Tipoc; xov A, 6 A Tipoc; xov B, f\ Be [iovac; xov A ^.expei xaxa xac; ev aux<2 jiovdSac xal 6 A apa xov B ^.expeT xaxa xac; ev auxw ^.ovdBac 6 A apa eauxov TioXXaTiXaaidaac; xupov xov B TieTiolrjxev. eav Be dpid^toc; eauxov TioXXaTiXaaidaac; xupov Tioifj, xal auxoc xupoc; eaxai. xupoc; apa xal 6 A- ouep oux UTioxeixai. oux apa 6 A xupoc; eaxlv. o^tolwc; 8f] Bel^o^tev, oxi ouB' aXXoc; ouBelc; xupoc; eaxi X W P^ x °u xexdpxou arco xfjc; ^tovdBoc; xal iSv Buo SiaXemovxcov ouep eSsri Belial. ia'. 'Eav dno jiovdSoc; ottoooiouv dpi'd^ol e^fjc; dvdXoyov Sow, 6 eXdxxwv xov jiel^ova ^icxpeT xaxa xiva xcov unap/ovx- cov ev xolc; dvdXoyov apti^ou;. A' 1 B r A E 'Eaxwaav diio (jiovdSoc; xfjc; A oTioaoiouv dpid^iol ec;fjc; dvdXoyov ol B, T, A, E- Xey«, oxi x<2v B, T, A, E 6 eXdxiaxoc; 6 B xov E ^lexpeT xaxa xiva xwv T, A. 'End yap eaxiv wc; f) A ^.ovac; Tipoc; xov B, ouxwc; 6 A Tipoc; xov E, ladxic; apa f] A [lovac; xov B dpid^tov ^.expeT xal 6 A xov E- evaXXdJ; apa ladxic; f) A ^lovac; xov A ^texpeT xal 6 B xov E. f) Be A jiovdc; xov A jiexpe'i xaxa xac; ev [Prop. 8.26]. And B is square. Thus, A is also square. The very opposite thing was assumed. C is thus not square. So, similarly, we can show that no other (number is) square either, apart from the third from the unit, and (all) those (numbers after that) which leave an interval of one (number). And so let A not be cube. I say that no other (num- ber) will be cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers) . For, if possible, let D be cube. And C is also cube [Prop. 9.8]. For it is the fourth (number) from the unit. And as C is to D, (so) B (is) to C. And B thus has to C the ratio which (some) cube (number has) to (some other) cube (number). And C is cube. Thus, B is also cube [Props. 7.13, 8.25]. And since as the unit is to A, (so) A (is) to B, and the unit measures A accord- ing to the units in it, A thus also measures B according to the units in (A). Thus, A has made the cube (num- ber) B (by) multiplying itself. And if a number makes a cube (number by) multiplying itself then it itself will be cube [Prop. 9.6]. Thus, A (is) also cube. The very oppo- site thing was assumed. Thus, D is not cube. So, simi- larly, we can show that no other (number) is cube either, apart from the fourth from the unit, and (all) those (num- bers after that) which leave an interval of two (numbers). (Which is) the very thing it was required to show. Proposition 11 If any multitude whatsoever of numbers is continu- ously proportional, (starting) from a unit, then a lesser (number) measures a greater according to some existing (number) among the proportional numbers. A' ' B 1 O 1 D 1 Let any multitude whatsoever of numbers, B, C, D, E, be continuously proportional, (starting) from the unit A. I say that, for B, C, D, E, the least (number), B, measures E according to some (one) of C, D. For since as the unit A is to B, so D (is) to E, the unit A thus measures the number B the same number of times as D (measures) E. Thus, alternately, the unit A 261 ETOIXEIfiN fl'. ELEMENTS BOOK 9 auxo) (iovdBac xal 6 B apa xov E [iexpei xaxd xac; ev xfi A ^iovd8a<;- <2>axe 6 eXdaaov 6 B xov ^.d^ova xov E ^e- xpeT xaxd xiva dpid^iov xwv UTtapxovxtov ev tou; dvdXoyov dpid^iou;. I16pia[Jia. Kal cpavepov, oxi rjv £)(£i xd^iv 6 (iexpwv aTto ^ovd8o<;, x/]v auxiqv e)(ei xal 6 xocd' 6v [iexpel arco xou (iexpou^evou em xo Tipo auxou. onep eSsi Sei^ai. LP'. 'Edv dno [iovd8o<; onoaoiouv dpid^iol e^fjc dvdXoyov Goiv, ucp' oacov dv 6 eayoLxoz uptoxwv dpidjiaiv ^.expfjxat,, utio xwv auxwv xal 6 Tiapa xfjv ^ovd8a ^.£xpr]Tf)r]aexai. A i ' Ei 1 B i 1 Z i 1 r i 1 Hi 1 A i 1 ©i 1 'Eaxwaav duo ^iovdBo<; OTCoaoiBrjTtoxouv dpid^tol dvdXoy- ov oi A, B, r, A' Xsyw, oxi ucp' oouv dv 6 A upcoxcov dpn3jji£5v jisxpfjxai, utio xwv auxcov xal 6 A [lexprydrpsxa!.. Mexpelo'dw yap 6 A utio xivoc Ttpwxou dpii&jiou xou E' Xeyw, oxi 6 E xov A [isxpel. \±r\ yap' xai eoxiv 6 E Tcpwxog, aTca<; 8e Ttp£>xo<; dpi'djioc; Ttpoc; auavxa, ov \Lr\ [LSTpei, Tipwxoc; eaxiv oi E, A apa Ttpoxoi Ttpoc dXXiqXouc eiaiv. xal stceI 6 E xov A [isxpei, ^.expeixw auxov xaxd xov Z' 6 E apa xov Z TtoXXaTtXaaidaag xov A TtSTtoirjXEv. TtdXiv, snel 6 A xov A [isxpsl xaxd xdc; ev xw T jiovdBac;, 6 A apa xov T TtoXXaTiXaaidaac xov A TieTcoirjXEv. dXXa \±r\v xal 6 E xov Z TtoXXaitXaaidaac; xov A TC£Tio[/]xev 6 apa tx xGv A, T Xaoz sraxl iw ex x£>v E, Z. saxiv apa 6? 6 A Ttpoc; xov E, 6 Z Tcp6<; xov T. oi 8e A, E TtpQxoi, oi 8e Tcpwxoi xal sXd)(iaxoi, oi 8s eXd)(iaxoi ^lexpouai xouc; xov auxov Xoyov s/ovxag iadxu; o xe f]you\L£voz xov f]you^ievov xal 6 stc6^svo<; xov STc6[ievov (jtexpsT apa 6 E xov T. jiexpsixw auxov xaxd xov H- 6 E apa xov H TtoXXaTtXaaidaag xov T TtETKHirpcev. dXXa (i/)v 8id xo Tipo xouxou xal 6 A xov B TioXXaTtXaaidaac; xov r TCETioirjxev. 6 apa ex xwv A, B laoc, saxl xco ex x£5v E, H. saxiv apa tbc; 6 A Ttpoc; xov E, 6 H Ttpoc; xov B. oi 8e A, E TcpGxoi, oi 8s Ttpwxoi xal eXd)(ioxoL, oi 8s eXd)(iaxoi dpi%ol measures D the same number of times as B (measures) E [Prop. 7.15]. And the unit A measures D according to the units in it. Thus, B also measures E according to the units in D. Hence, the lesser (number) B measures the greater E according to some existing number among the proportional numbers (namely, D). Corollary And (it is) clear that what(ever relative) place the measuring (number) has from the unit, the (number) according to which it measures has the same (relative) place from the measured (number), in (the direction of the number) before it. (Which is) the very thing it was required to show. Proposition 12 If any multitude whatsoever of numbers is continu- ously proportional, (starting) from a unit, then however many prime numbers the last (number) is measured by, the (number) next to the unit will also be measured by the same (prime numbers). A i 1 E i 1 B i 1 F i 1 C i 1 G' 1 D 1 Hi 1 Let any multitude whatsoever of numbers, A, B, C, D, be (continuously) proportional, (starting) from a unit. I say that however many prime numbers D is measured by, A will also be measured by the same (prime numbers). For let D be measured by some prime number E. I say that E measures A. For (suppose it does) not. E is prime, and every prime number is prime to every num- ber which it does not measure [Prop. 7.29]. Thus, E and A are prime to one another. And since E measures D, let it measure it according to F. Thus, E has made D (by) multiplying F. Again, since A measures D ac- cording to the units in C [Prop. 9.11 corr.], A has thus made D (by) multiplying C. But, in fact, E has also made D (by) multiplying F. Thus, the (number cre- ated) from (multiplying) A, C is equal to the (number created) from (multiplying) E, F. Thus, as A is to E, (so) F (is) to C [Prop. 7.19]. And A and E (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ra- tio as them) [Prop. 7.21], and the least (numbers) mea- sure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the lead- 262 ETOIXEIfiN fl'. ELEMENTS BOOK 9 [isxpouai xou<; xov auxov Xoyov £)(ovxac; auxolc; iadxig 5 xe f)you|ievo<; xov fjyou^ievov xal 6 eit6\±evoq xov feraD^isvov ^expei apa 6 E xov B. jiexpeixco auxov xaxd xov 9- 6 E apa xov O TioXXarcXaaidaat; xov B KETioirjxev. dXXd jir]v xal 6 A sauxov 7ioXXanXaaidaa<; xov B 7t£7toiT)xev 6 apa ex xwv E, 9 iao<; eaxl xo duo xou A. eaxiv apa d><; 6 E npoc, xov A, 6 A ■rcpoc xov 0. oi Se A, E Ttpwxoi, oi 8e TtpGxoi xal eXd)(iaxoi, oi 8s sXdxiaxoi (iexpouai xouc; xov auxov Xoyov £)(ovxac; iadxig 6 fjyou^evog xov fjyoujisvov xal 6 sk6\xsmoc, xov eho^isvov [iexpsl apa 6 E xov A w<; r\\o\j\ievoz fjyoujievov. dXXd [i/]v xal ou (Jtexpel - onsp dBuvaxov. oux apa oi E, A Tipwxoi npbz dXXrjXouc; eioiv. auvdexoi apa. oi 8s auvdexoi uno [upcbxou] dpii9(io0 xivog ^texpouvxai. xal snsl 6 E upaixoc; unoxeixai, 6 6e upwxoc; Otto exepou apiS^ou ou [isxpelxai fj ucp' eauxou, 6 E apa xou? A, E ^isxpeT- uoxe 6 E xov A ^lexpeT. ^lexpei 8s xal xov A - 6 E apa xou<; A, A ^.expeT. 6^io[«<; 8f) Ssi^o^iev, oxi ucp' oaov dv 6 A upwxov dpid^iGv [isxprjxai, uuo iSv auxCSv xal 6 A [isxprydrjaexai- orcep e8si 8ei<;ai. ing, and the following the following [Prop. 7.20] . Thus, E measures C. Let it measure it according to G. Thus, E has made C (by) multiplying G. But, in fact, via the (proposition) before this, A has also made C (by) multi- plying B [Prop. 9.11 corr.]. Thus, the (number created) from (multiplying) A, B is equal to the (number created) from (multiplying) E, G. Thus, as A is to E, (so) G (is) to B [Prop. 7.19]. And A and E (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures B. Let it measure it according to H. Thus, E has made B (by) multiplying H. But, in fact, A has also made B (by) multiplying itself [Prop. 9.8]. Thus, the (number created) from (multiplying) E, H is equal to the (square) on A. Thus, as E is to A, (so) A (is) to H [Prop. 7.19]. And A and E are prime (to one an- other), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures A, as the leading (measuring the) leading. But, in fact, (E) also does not measure (A). The very thing (is) impossible. Thus, E and A are not prime to one another. Thus, (they are) composite (to one another). And (numbers) composite (to one another) are (both) measured by some [prime] number [Def. 7.14]. And since E is assumed (to be) prime, and a prime (number) is not measured by another number (other) than itself [Def. 7.11], E thus measures (both) A and E. Hence, E measures A. And it also measures D. Thus, E measures (both) A and D. So, similarly, we can show that however many prime numbers D is measured by, A will also be measured by the same (prime numbers). (Which is) the very thing it was required to show. 'Edv dno [iovd8o<; 6noooiouv dpid^iol tEjfi dvdXoyov Goiv, 6 5e \xeto. xr]v ^iovd8a npfiixoc; fj, 6 [leyiaxog un ouSevoc; [aXXou] jjiexpiq'drpsxai napsE, x«v UTtap/ovxiov ev xou; dvdXoyov dpid^iou;. 'Eaxwaav and ^lovdSoc; oiioaoiouv apidfjioi e^rj<; dvdXoyov oi A, B, T, A, 6 8e ^texd x/]v ^tovd8a 6 A TtpGxoc; eaxw Xcyw, oxi 6 ^tsyiaxoc; auxfiv 6 A bn ouSevoc; aXXou [jle- xp/]i9r)Gexai Ttapec; x«v A, B, T. Proposition 13 If any multitude whatsoever of numbers is continu- ously proportional, (starting) from a unit, and the (num- ber) after the unit is prime, then the greatest (number) will be measured by no [other] (numbers) except (num- bers) existing among the proportional numbers. Let any multitude whatsoever of numbers, A, B, C, D, be continuously proportional, (starting) from a unit. And let the (number) after the unit, A, be prime. I say 263 ETOIXEIfiN fl'. ELEMENTS BOOK 9 B r h Ah Hi ' ©i ' El yap 8uvax6v, [isxpeio'dw utio tou E, xal 6 E ^ir)8evl xfiv A, B, r eaxw 6 auxoc;. cpavepov 8rj, oxi 6 E Tiptoxoc; oux eaxiv. el yap 6 E Tipfixoc; eaxi xal (jiexpel xov A, xal xov A [i£Tpr\o£i Tipaixov ovxa [if] uv auxai 6 auxoc;' ouep eaxlv d8uvaxov. oux apa 6 E Tipfixoc; eaxiv. auvdexoc; apa. Tide; 8e auvdexoc; dpidjioc; utio npwxou xivoc; dpidjiou [iexpeTxar 6 E apa utio Tipwxou xivoc; dpidjiou (icxpelxai. Xeyio 8r|, oxi On' ouSevoc; aXXou Tiptoxou [icxprydriaexai tiX/jv xou A. el yap Ocp° exepou [Jiexpelxai 6 E, 6 8s E xov A [iexpel, xdxelvoc; apa xov A [L£Tpr\oev a>axe xal xov A (iexpr]aei Tipaixov ovxa \ir\ &>v auxfi 6 auxoc;' onep eaxlv d8uvaxov. 6 A apa xov E (jiexpel. xal inz\ 6 E xov A [iexpel, [lexpeixw auxov xaxd xov Z. Xeyco, oxi 6 Z ouSevl tuv A, B, T eaxiv 6 auxoc;. el yap 6 Z evl xaiv A, B, T eaxiv 6 auxoc; xal [iexpel xov A xaxd xov E, xal elg apa xwv A, B, T xov A [iexpel xaxd xov E. dXXa sic, xwv A, B, r xov A [iexpel xaxd xiva xwv A, B, T- xal 6 E apa evl xwv A, B, T eaxiv 6 auxoc;' onep oux UTioxeixai. oux apa 6 Z evl xwv A, B, T eaxiv 6 auxoc;. 6[ioia>c; Br| 8eii;o[i£v, oxi [lexpelxai 6 Z utio xou A, Beixvuvxec; TidXiv, oxi 6 Z oux eaxi Tipcoxoc;. ei yap, xal [iexpel xov A, xal xov A [iexpr|aei upSSxov ovxa \xf\ d>v auxcS 6 auxoc;' oicep eaxlv dSuvaxov oux apa itpfixoc; eaxiv 6 Z' auvdexoc; apa. dnac; 8e auvdexoc; dpi'djioc; utio Ttpwxou xivoc; dpid[iou (lexpelxai' 6 Z apa utio upcoxou xivoc; dpi-djiou [lexpelxai. Xeyco 8rj, oxi ucp' exepou jcpwxou ou [i£xp/]iL>r]aexai tiX/]v xou A. ei yap exepoc; xic; Tipwxoc; xov Z (iexpel, 6 8e Z xov A jiexpel, xdxelvoc; apa xov A [iexpf|aei' waxe xal xov A [lexprpei Tipwxov ovxa \ir\ <2>v auxfi) 6 auxoc;' oTiep eaxlv dBuvaxov. 6 A apa xov Z [iexpel. xal end 6 E xov A [iexpel xaxd xov Z, 6 E apa xov Z TtoXXauXaaidaac; xov A TiCTioirjxev. dXXa [irjv xal 6 A xov r TioXXaTiXaaidaac; xov A TiCTioirjxev 6 apa ex xwv A, T Xaoc, eaxi to ex xCSv E, Z. dvdXoyov apa eaxlv wc; 6 A Tipoc; xov E, ouxcoc; 6 Z Tipoc; xov T. 6 8e A xov E [iexpel' xal 6 Z apa xov r [iexpel. [lexpeixw auxov xaxd xov H. o^ioiwc; 8r) 8eic;o[iev, oxi 6 H ou8evl xwv A, B eaxiv 6 auxoc;, xal oxi [lexpelxai utio xou A. xal enel 6 Z xov V [iexpel xaxd xov H, 6 Z apa xov H TioXXauXaaidaac; xov V 7ieiio(r]xev. dXXa \ir\\> xal 6 A xov B TioXXanXaaidaac; xov T 7i£7iolr)xev 6 apa ex xov A, B laoc; eaxl xw ex xwv Z, H. dvdXoyov apa <!><; 6 A Tipoc; xov Z, 6 H Tipoc; xov B. [xexpeT Be 6 A xov Z- (iexpel apa xal 6 H xov B. (xexpeixw auxov xaxd xov 0. 6(ioiw<; 8r) 8eic;o[iev, oxi 6 9 ifi A oux eaxiv 6 auxoc;. xal cticI 6 H xov that the greatest of them, D, will be measured by no other (numbers) except A, B, C. A' 1 E i 1 F ^ For, if possible, let it be measured by E, and let E not be the same as one of A, B, C. So it is clear that E is not prime. For if E is prime, and measures D, then it will also measure A, (despite ^4) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, E is not prime. Thus, (it is) composite. And every composite number is measured by some prime number [Prop. 7.31]. Thus, E is measured by some prime num- ber. So I say that it will be measured by no other prime number than A. For if E is measured by another (prime number), and E measures D, then this (prime number) will thus also measure D. Hence, it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, A mea- sures E. And since E measures D, let it measure it ac- cording to F. I say that F is not the same as one of A, B, C. For if F is the same as one of A, B, C, and measures D according to E, then one of A, B, C thus also measures D according to E. But one of A, B, C (only) measures D according to some (one) of A, B, C [Prop. 9.11]. And thus E is the same as one of A, B, C. The very oppo- site thing was assumed. Thus, F is not the same as one of A, B, C. Similarly, we can show that F is measured by A, (by) again showing that F is not prime. For if (F is prime), and measures D, then it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, F is not prime. Thus, (it is) composite. And every composite number is measured by some prime number [Prop. 7.31]. Thus, F is measured by some prime number. So I say that it will be measured by no other prime number than A. For if some other prime (number) measures F, and F measures D, then this (prime number) will thus also measure D. Hence, it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12]. The very thing is impossible. Thus, A measures F. And since E measures D according to F, E has thus made D (by) multiplying F. But, in fact, A has also made D (by) multiplying C [Prop. 9.11 corr.]. Thus, the (number created) from (multiplying) A, C is equal to the (number created) from (multiplying) E, F. Thus, proportionally, as A is to E, so F (is) to C [Prop. 7.19]. And A measures 264 ETOIXEIfiN fl'. ELEMENTS BOOK 9 B jjiexpsT xaxd xov 0, 6 H dpa xov TtoXXaTiXaa(.daa<; xov B 7i£iio(r]xev. dXXd ^f)V xal 6 A eauxov TioXXaTcXaaidaac; xov B TiETtoirjxev 6 dpa utio 6, H iaoz eaxl iu dTio xou A Teipaywvw' eaxiv dpa u<;6 8 Tipoc; xov A, 6 A Tip6<; tov H. [icxpeT 8e 6 A xov H- [icxpeT dpa xal 6 O xov A npoxov ovxa \ir\ &v auxSS 6 auxoc oTiep dxoTiov. oux dpa 6 ^eyiaxoc; 6 A utio exepou dpidjiou ueTpryft^asxai Tiape^ x«v A, B, T- ojiep sSei 5sT5ai. E. Thus, F also measures C. Let it measure it according to G. So, similarly, we can show that G is not the same as one of A, B, and that it is measured by A. And since F measures C according to G, F has thus made C (by) multiplying G. But, in fact, A has also made C (by) mul- tiplying B [Prop. 9.11 corr.]. Thus, the (number created) from (multiplying) A, B is equal to the (number created) from (multiplying) F, G. Thus, proportionally, as A (is) to F, so G (is) to B [Prop. 7.19]. And A measures F. Thus, G also measures B. Let it measure it according to H. So, similarly, we can show that H is not the same as A. And since G measures B according to H, G has thus made B (by) multiplying H. But, in fact, A has also made B (by) multiplying itself [Prop. 9.8]. Thus, the (number created) from (multiplying) H, G is equal to the square on A. Thus, as H is to A, (so) A (is) to G [Prop. 7.19]. And A measures G. Thus, H also measures A, (despite A) being prime (and) not being the same as it. The very thing (is) absurd. Thus, the greatest (number) D cannot be measured by another (number) except (one of) A, B, C. (Which is) the very thing it was required to show. 18'. 'Edv skctyioToz dpidjioe; utio icpwxwv dpid^wv ^.expfjxai, bn ou8evo<; dXXou icpwxou dpid^ou jiexp/jOiqaexai Tiape?; xwv 15 dpxfjt; ^lexpouvxtov. B ^ r h Ah 2' — i 'EXd)(ioxoc; yap dpn&jioc; 6 A utio icpoxcov dpidjifiv xwv B, T, A ^expeicrdw Xeyw, oxi 6 A Ok' ouSevcx; dXXou itpwxou dpiiSjioO (lexprydrpexai uape^ xaiv B, T, A. EE yap Suvaxov, (lexpeio-dio utio Tipcoxou xou E, xal 6 E [i.r)8evl xSv B, T, A eaxco 6 auxog. xal fenel 6 E xov A [iexpeT, (icxpeixio auxov xaxa xov Z' 6 E dpa xov Z TtoXXaTiXaaidaag xov A icsitoi/jxev. xal jiexpelxai 6 A utio Ttpcjxcov dpii9[i£iv xcov B, T, A. edv Be Buo dpid^iol tioX- XaTcXaoidaavxs<; dXXf]Xou<; koiGSoi xiva, xov Be yevo^ievov ec; auxwv fiexpfj xlc Ttptoxoc api'&y.oz, xal eva xwv fE, dp^fjc; \iSTprpei' o\ B, T, A dpa eva xwv E, Z [lexprpouaiv. xov ^xev ouv E ou ^lexpiqaoumv 6 yap E Tcpox6<; eaxi xal ouBevl xwv B, T, A 6 auxo<;. xov Z dpa ^texpouaiv eXdaaova ovxa xou A' oTtep dBuvaxov. 6 yap A UTioxeixai eXd)(io"xoc utio xwv B, T, A ^.expou^tevot;. oux dpa xov A \iSTpr\ae\. TipGxoc; dpid^icx; Tiape^ xwv B, T, A- OTiep eSei BeT^ai. Proposition 14 If a least number is measured by (some) prime num- bers then it will not be measured by any other prime number except (one of) the original measuring (num- bers). A 1 B i 1 F h Dh H For let A be the least number measured by the prime numbers B, C, D. I say that A will not be measured by any other prime number except (one of) B, C, D. For, if possible, let it be measured by the prime (num- ber) E. And let E not be the same as one of B, C, D. And since E measures A, let it measure it according to F. Thus, E has made A (by) multiplying F. And A is mea- sured by the prime numbers B, C, D. And if two num- bers make some (number by) multiplying one another, and some prime number measures the number created from them, then (the prime number) will also measure one of the original (numbers) [Prop. 7.30]. Thus, B, C, D will measure one of E, F. In fact, they do not measure E. For E is prime, and not the same as one of B, C, D. Thus, they (all) measure F, which is less than A. The very thing (is) impossible. For A was assumed (to be) the least (number) measured by B, C, D. Thus, no prime 265 ETOIXEIfiN fl'. ELEMENTS BOOK 9 is'. 'Edv xpeTc; dpid^toi ec;rj<; dvaXoyov Saiv eXd)(iaxoi xwv xov auxov Xoyov e)(6vxwv auxolc;, Buo ottoioiouv auv- xcdevxec; Tcpoc xov Xomov upwxoi Eiaiv. A E 2 A i 1 i 1 1 B i 1 r i 1 'Eaxwaav xpeu; apid^oi i^f\z dvaXoyov eXd)(iaxoi xSv xov auxov Xoyov e)(6vxwv auxolc; oi A, B, E Xeyw, oxi xwv A, B, T 5uo otioioioDv auvxcdevxeg npbc, xov Xomov upwxoi eiaiv, oi [lev A, B Tipot; xov T, oi 8s B, T npbc, xov A xai exi oi A, r 7tp6<; xov B. EiXr ! )(p , dwaav yap eXd/iaxoi apidjjioi xwv xov auxov Xoyov e^ovxwv xo ^ -A- 1 B, r Buo oi AE, EZ. cpavepov 8r), oxi 6 (isv AE eauxov TioXXajcXaaidaac; xov A TteTcoirjxev, xov Be EZ TioXXaicXaaidaac; xov B iieTcoir)xev, xai ixi 6 EZ eauxov TtoXXajcXaaidaac; xov T KSTioirjXEv. xai eicei oi AE, EZ eXdxiaxoL eiaiv, itpwxoi npoc; dXXr|Xou<; eiaiv. eav Be Buo dpi%oi itpwxoi upog dXXrjXouc; waiv, xai auva|icp6xepoc; npoc exdxepov Ttpwxoc; eaxiv xai 6 AZ dpa upoc; exdxepov xwv AE, EZ npwxoc; eaxiv. dXXd xai 6 AE npbc, xov EZ itpwxoc; eaxiv oi AZ, AE dpa npoc; xov EZ Ttpwxoi eiaiv. eav Be Buo dpidjioi upog xiva dpL^jiov itpwxoi waiv, xai 6 zl, auxwv yevo^xevoc; 7cp6c; xov Xomov itpwxoc; eaxiv waxe 6 ex xwv ZA, AE npoc; xov EZ Ttpwxoc; eaxiv waxe xai 6 ex xwv ZA, AE icpot; xov arco xou EZ Tcpwxoc; eaxiv. [eav yap Buo dpii9(ioi Ttpwxoi Ttpoc; dXXr]Xouc; waiv, 6 ex xou tvbc, auxwv yevo^ievoc; Ttpoc; xov Xoitiov Ttpwxoc; eaxiv]. dXX' 6 ex xwv ZA, AE 6 dno xou AE eaxi jiexa xou ex xwv AE, EZ- 6 dpa duo xou AE jiexa xou ex xwv AE, EZ Ttpoc; xov aTio xou EZ Ttpwxoc; eaxiv. xai eaxiv 6 uev duo xou AE 6 A, 6 Be ex xwv AE, EZ 6 B, 6 Be duo xou EZ 6 E oi A, B dpa auvxei!)evxe<; Ttpoc; xov T Ttpwxoi eiaiv. ojioiwc; Srj 8eic;ouev, oxi xai oi B, T npbc, xov A Ttpwxoi eiaiv. Xeyw Srj, oxi xai oi A, T npbc, xov B Ttpwxoi eiaiv. CTtei yap 6 AZ Ttpoc; exdxepov xwv AE, EZ Ttpwxoc eaxiv, xai 6 aTto xou AZ npbc, xov ex xwv AE, EZ Ttpwxoc eaxiv. dXXd xw drco xou AZ i'aoi eiaiv oi aTto xwv AE, EZ jiexa xou Bic ex xwv AE, EZ- xai oi aTto xwv AE, EZ dpa (icxa xou Si? bnb xwv AE, EZ Ttpoc xov (mo xwv AE, EZ Ttpwxoi [eiai]. BieXovxi oi duo xwv AE, EZ ^icxa xou aTtac; bnb AE, EZ Ttpoc xov Oito AE, EZ Ttpwxoi eiaiv. exi BieXovxi oi aTto xwv AE, EZ dpa Ttpoc xov utio AE, EZ Ttpwxoi eiaiv. xai eaxiv 6 jxev number can measure A except (one of) B, C, D. (Which is) the very thing it was required to show. Proposition 15 If three continuously proportional numbers are the least of those (numbers) having the same ratio as them then two (of them) added together in any way are prime to the remaining (one) . D E F A i 1 i 1 1 Bi ' C' ' Let A, B, C be three continuously proportional num- bers (which are) the least of those (numbers) having the same ratio as them. I say that two of A, B, C added to- gether in any way are prime to the remaining (one), (that is) A and B (prime) to C, B and C to A, and, further, A and C to B. Let the two least numbers, DE and EF, having the same ratio as A, B, C, have been taken [Prop. 8.2]. So it is clear that DE has made A (by) multiplying it- self, and has made B (by) multiplying EF, and, fur- ther, EF has made C (by) multiplying itself [Prop. 8.2]. And since DE, EF are the least (of those numbers hav- ing the same ratio as them), they are prime to one an- other [Prop. 7.22]. And if two numbers are prime to one another then the sum (of them) is also prime to each [Prop. 7.28]. Thus, DF is also prime to each of DE, EF. But, in fact, DE is also prime to EF. Thus, DF, DE are (both) prime to EF. And if two numbers are (both) prime to some number then the (number) created from (multiplying) them is also prime to the remaining (num- ber) [Prop. 7.24]. Hence, the (number created) from (multiplying) FD, DE is prime to EF. Hence, the (num- ber created) from (multiplying) FD, DE is also prime to the (square) on EF [Prop. 7.25]. [For if two num- bers are prime to one another then the (number) created from (squaring) one of them is prime to the remaining (number).] But the (number created) from (multiplying) FD, DE is the (square) on DE plus the (number cre- ated) from (multiplying) DE, EF [Prop. 2.3]. Thus, the (square) on DE plus the (number created) from (multi- plying) DE, EF is prime to the (square) on EF. And the (square) on DE is A, and the (number created) from (multiplying) DE, EF (is) B, and the (square) on EF (is) C. Thus, A, B summed is prime to C. So, similarly, we can show that B, C (summed) is also prime to A. So I say that A, C (summed) is also prime to B. For since 266 ETOIXEIfiN fl'. ELEMENTS BOOK 9 duo xoO AE 6 A, 6 Be: utio xwv AE, EZ 6 B, 6 8s dtKo xou DF is prime to each of DE, EF then the (square) on DF EZ 6 T. ol A, T dpa auvxe'devxec; Tipoc xov B Ttpfixoi elcnv is also prime to the (number created) from (multiplying) oTisp eBsi hzXiai. DE, EF [Prop. 7.25]. But, the (sum of the squares) on DE, EF plus twice the (number created) from (multiply- ing) DE, EF is equal to the (square) on DF [Prop. 2.4]. And thus the (sum of the squares) on DE, EF plus twice the (rectangle contained) by DE, EF [is] prime to the (rectangle contained) by DE, EF. By separation, the (sum of the squares) on DE, EF plus once the (rect- angle contained) by DE, EF is prime to the (rectangle contained) by DE, EFJ Again, by separation, the (sum of the squares) on DE, EF is prime to the (rectangle contained) by DE, EF. And the (square) on DE is A, and the (rectangle contained) by DE, EF (is) B, and the (square) on EF (is) C. Thus, A, C summed is prime to B. (Which is) the very thing it was required to show. t Since if a f3 measures a 2 + (3 2 + 2 a (3 then it also measures a 2 + ffi + a (3, and vice versa. If'. 'Edv Buo dpid|jiol Ttpwxoi. 7ip6<; dXXr|Xouc Soiv, oux eaxai cbc; 6 npcbxoc; Ttpoc; xov Bsuxspov, ouxtoc; 6 Beuxepoe; 7ip6<; aXXov xivd. A i ' B i ' r i 1 Auo yap dpu^oi oi A, B nptoxoi npoc; dXXr|Xouc; eoxto- aocv Xeyw, oxi oux eaxiv cbc; 6 A npoc, xov B, ouxw<; 6 B 7ip6<; aXXov xivd. El yap Buvaxov, eaxco cbc; 6 A 7ip6<; xov B, 6 B Ttpoe; xov r. ol Be A, B upOxoi, oi Be Ttpcbxoi xal eXd)(iaxoi, oi Be sXa/iaxoi dpidjiol ^.expoOai xouc; xov auxov Xoyov exovxag iadxic; 8 xe f}yo'j\±evoz xov f]Y ou ^ £vov xal ° etco^isvoc, xov £ttojj£vov (jiexpsl dpa 6 A xov B cbc, r)you[jievoc; fiyoujievov. (iexpeT Be xal eauxov 6 A dpa xou<; A, B ^lexpei itpcbxouc; ovxac; 7ip6<; dXXfjXouc ouep dxorcov. oux apa saxai cbc; 6 A npoc, xov B, ouxtoc; 6 B Ttpoc; xov T- 6mp eSei Belc^ai. 'Edv Soiv oaoiBrjTioxouv dpid^toi e^rjc; dvdXoyov, oi Be axpoi auxcbv upwxoi rcpoc; dXXrjXouc; Saiv, oux eaxai cbc; 6 upCSxoc; Tipoc; xov Beruxepov, ouxck 6 eo)(axo<; Ttpoc; dXXov Proposition 16 If two numbers are prime to one another then as the first is to the second, so the second (will) not (be) to some other (number) . A' 1 B 1 C' ' For let the two numbers A and B be prime to one another. I say that as A is to B, so B is not to some other (number). For, if possible, let it be that as A (is) to B, (so) B (is) to C. And A and B (are) prime (to one an- other). And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the lead- ing, and the following the following [Prop. 7.20] . Thus, A measures B, as the leading (measuring) the leading. And (A) also measures itself. Thus, A measures A and B, which are prime to one another. The very thing (is) ab- surd. Thus, as A (is) to B, so B cannot be to C. (Which is) the very thing it was required to show. Proposition 17 If any multitude whatsoever of numbers is continu- ously proportional, and the outermost of them are prime to one another, then as the first (is) to the second, so the 267 ETOIXEIfiN fl'. ELEMENTS BOOK 9 xivd. 'Eaxwaav 6aoi8r]7ioTouv api'd^oi e^fj<; dvdXoyov oi A, B, T, A, ol 8e dxpoi auxfiv oi A, A npoxoi npoc dXXr|Xouc; eaxcoaav Xeyco, §xi oux eaxiv cb<; 6 A upoc; xov B, ouxok 6 A rcpoc; dXXov xivd. A' 1 B ' ' r ' 1 A i ' Ei ' El yap 8uvax6v, eaxco foe. 6 A npoz xov B, ouxox 6 A 7ip6<; xov E- evaXXd?; dpa eaxiv 6? 6 A Ttpoc xov A, 6 B Ttpoc; xov E. oi 8e A, A itpcoxoi, oi 8e TtpGxoi xai tXayiaxoi, oi 8s eXd)(iaxoi apidjiol ^texpouai xou<; xov auxov Xoyov e)(ovxa<; iadxi<; o xe f]you^ievo<; TOV /]You^evov xa ' 1 ° i^o[ievoc, xov CTtojievov. [jiexpei dpa 6 A xov B. xai eaxiv foe, 6 A npbe xov B, 6 B icpoc^ xov T. xai 6 B dpa xov T jxexpeT - &axe xai 6 A xov T jiexpei. xai inei eaxiv (be; 6 B npog xov T, 6 r Kpoc; xov A, ^.expeT Be 6 B xov T, ^.expeT dpa xai 6 T xov A. dXX' 6 A xov T E^expei' waxe 6 A xai xov A ^expert. ^expeT 8s xai eauxov. 6 A dpa xou<; A, A ^expeT n:pa>xou<; ovxa<; 7ip6<; dXXf|Xou<; - oiiep eaxiv dBuvaxov. oux dpa eaxai foe, 6 A 7ip6<; xov B, oux«<; 6 A npbe dXXov xivd- ouep e8ei 8eT?ai. IT]'. Auo dpi%£5v Bo'devxwv emaxetjjaa'dai, el Buvaxov eaxiv auxolg xpixov dvdXoyov rcpoaeupeTv. a i 1 r i 1 B i 1 A i 1 "Eaxwaav oi BoiJevxet; 8uo dprdjioi oi A, B, xai 8eov eaxw eraaxe^aa'dai, si 8uvax6v eaxiv auxolc; xpixov dvdXoyov Ttpoaeupefv. Oi 8r) A, B ffzoi Tipfixoi upog dXXr]Xouc; eiaiv fj ou. xai si Tipwxoi npbe. dXXf]Xouc; eiaiv, SeBeixxai, oxi dBuvaxov eaxiv auxou; xpixov dvdXoyov TtpoaeupeTv. AXXd 8r) \±r\ eaxtoaav oi A, B rcpfiixoi npbe dXXrjXouc;, xai 6 B eauxov TioXXanXaaidaag xov T noieixw. 6 A §r| xov T f]xoi jiexpei f] ou jiexpei. |iexpeix« npoxepov xaxd xov A- 6 A dpa xov A TroXXanXaaidaac; xov T nenoi/jxev. dXXa ^ir)v xai 6 B eauxov noXXaTiXaaidaag xov T nz%oir\xev b dpa last will not be to some other (number). Let A, B, C, D be any multitude whatsoever of con- tinuously proportional numbers. And let the outermost of them, A and D, be prime to one another. I say that as A is to B, so D (is) not to some other (number) . A' 1 B ' 1 Ci ' D' ' Ei ' For, if possible, let it be that as A (is) to B, so D (is) to E. Thus, alternately, as A is to D, (so) B (is) to E [Prop. 7.13]. And A and D are prime (to one another). And (numbers) prime (to one another are) also the least (of those numbers having the same ra- tio as them) [Prop. 7.21]. And the least numbers mea- sure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the lead- ing, and the following the following [Prop. 7.20] . Thus, A measures B. And as A is to B, (so) B (is) to C. Thus, B also measures C. And hence A measures C [Def. 7.20]. And since as B is to C, (so) C (is) to D, and B mea- sures C, C thus also measures D [Def. 7.20] . But, A was (found to be) measuring C. And hence A also measures D. And (A) also measures itself. Thus, A measures A and D, which are prime to one another. The very thing is impossible. Thus, as A (is) to B, so D cannot be to some other (number) . (Which is) the very thing it was required to show. Proposition 18 For two given numbers, to investigate whether it is possible to find a third (number) proportional to them. A i 1 C' ' B i 1 Di 1 Let A and B be the two given numbers. And let it be required to investigate whether it is possible to find a third (number) proportional to them. So A and B are either prime to one another, or not. And if they are prime to one another then it has (already) been show that it is impossible to find a third (number) proportional to them [Prop. 9.16]. And so let A and B not be prime to one another. And let B make C (by) multiplying itself. So A either mea- sures, or does not measure, C. Let it first of all measure (C) according to D. Thus, A has made C (by) multiply- 268 ETOIXEIfiN fl'. ELEMENTS BOOK 9 ex xfiv A, A iaoc; eaxi xfi dmo xou B. eaxiv apa &>z 6 A 7ip6<; xov B, 6 B Tipoc; xov A- xdig A, B apa xpixoc; dpid^toc; dvdXoyov Tipoar]upr)xai 6 A. AXXa 8f] [i/j [icxpeixto 6 A xov E Xeyw, oxi xolc; A, B dSuvaxov eaxi xpixov dvdXoyov TtpoaeupeTv dpidjiov. et yap Buvaxov, n;poar]upr]a , d« 6 A. 6 apa ex xwv A, A Iaoc eaxi x£> duo xou B. 6 8e and xou B eaxiv 6 T' 6 apa ex xwv A, A iaoc eaxi xC) T. wote 6 A xov A TtoXXaitXaaidaac xov r 7te7tcu7]xev 6 A apa xov T ^texpeT xaxa xov A. dXXa (i/]v UTtoxeixai xal {jltj ^.expwv orcep axoitov. oux apa Buvaxov eaxi xoi"c A, B xpixov dvdXoyov Ttpoaeupelv dpid^ov, oxav 6 A xov T [jlt) nexpfj- oitep e8ei 8el5ai. 10'. Tpiwv dpnD^iwv BoiSevxMv eTiiaxe<j;aaiL>ai, iraxe Buvaxov eaxiv auxoic xexapxov dvdXoyov Ttpoaeupeiv. A' 1 B ' ' r ' ' A i 1 E' ' "Eaxwaav oi So-devxec xpeTc dpi'd^oi oi A, B, T, xal Seov eaxto £TuaxecJ>acrdai, Ttoxe Buvaxov eaxiv auxoic xexapxov dvdXoyov Ttpoaeupeiv. "Hxoi ouv oux eiaiv ec^fjc dvdXoyov, xal oi axpoi auxfiv Tipwxoi Tipoc dXXf|Xouc eiaiv, fj ec;fjc eiaiv dvdXoyov, xal oi axpoi auxwv oux eiai Kpwxoi Tipoc dXXr|Xouc, fj ouxe e<;rjc eiaiv dvdXoyov, ouxe oi axpoi auxwv Tipoxoi Tipoc dXXr|Xouc eiaiv, fj xal e^fjc eiaiv dvdXoyov, xal oi axpoi auxfiv TtpGxoi Tipoc dXXf]Xouc eiaiv. Ei (jiev ouv oi A, B, T ec;rjc eiaiv dvdXoyov, xal oi axpoi auxwv oi A, T upGxoi Tipoc dXXf|Xouc eiaiv, SeBeixxai, oxi dBuvaxov eaxiv auxoic xexapxov dvdXoyov Tipoaeupeiv dpid^iov. [XT] eaxoaav 8/] oi A, B, T ei;rjc dvdXoyov xwv dxpwv TtdXiv ovxwv Ttpwxtov Tipoc dXXf|Xouc. Xey«, oxi xal ouxwc dSuvaxov eaxiv auxoic xexapxov dvdXoyov Tipo- aeupeTv. ei yap 8uvax6v, Tipoaeuprp'dw 6 A, waxe elvai cbc xov A Tipoc xov B, xov r Tipoc xov A, xal yeyovexo 6? 6 B Tipoc xov r, 6 A Tipoc xov E. xal eitei eaxiv cbc [lev 6 A Tipoc xov B, 6 r Tipoc xov A, «c Be 6 B Tipoc xov T, 6 A Tipoc xov E, 8i' i'aou apa cbc 6 A Tipoc xov T, 6 T Tipoc xov E. oi Be A, T Ttpwxoi, oi Be TipGxoi xal eXd)(iaxoi, oi Be eXd)(iaxoi ing D. But, in fact, B has also made C (by) multiplying itself. Thus, the (number created) from (multiplying) A, D is equal to the (square) on B. Thus, as A is to B, (so) B (is) to D [Prop. 7.19]. Thus, a third number has been found proportional to A, B, (namely) D. And so let A not measure C. I say that it is impossi- ble to find a third number proportional to A, B. For, if possible, let it have been found, (and let it be) D. Thus, the (number created) from (multiplying) A, D is equal to the (square) on B [Prop. 7.19]. And the (square) on B is C. Thus, the (number created) from (multiplying) A, D is equal to C. Hence, A has made C (by) multiplying D. Thus, A measures C according to D. But (A) was, in fact, also assumed (to be) not measuring (C). The very thing (is) absurd. Thus, it is not possible to find a third number proportional to A, B when A does not measure C. (Which is) the very thing it was required to show. Proposition 19 f For three given numbers, to investigate when it is pos- sible to find a fourth (number) proportional to them. A' 1 B' 1 C' ' D' 1 Ei 1 Let A, B, C be the three given numbers. And let it be required to investigate when it is possible to find a fourth (number) proportional to them. In fact, (A, B, C) are either not continuously pro- portional and the outermost of them are prime to one another, or are continuously proportional and the outer- most of them are not prime to one another, or are neither continuously proportional nor are the outermost of them prime to one another, or are continuously proportional and the outermost of them are prime to one another. In fact, if A, B, C are continuously proportional, and the outermost of them, A and C, are prime to one an- other, (then) it has (already) been shown that it is im- possible to find a fourth number proportional to them [Prop. 9.17]. So let A, B, C not be continuously propor- tional, (with) the outermost of them again being prime to one another. I say that, in this case, it is also impossible to find a fourth (number) proportional to them. For, if possible, let it have been found, (and let it be) D. Hence, it will be that as A (is) to B, (so) C (is) to D. And let it be contrived that as B (is) to C, (so) D (is) to E. And since 269 ETOIXEIfiN fl'. ELEMENTS BOOK 9 [iexpouai xou<; tov auxov Xoyov e)(ovxac; 6 xe fjyoujievoc; xov rjyoujievov xal 6 CTio^evoc; xov euojievov. jiexpeT apa 6 A tov T tdc, f]you(jievoc; fjyou^evov. jiexpel 8e xai eauxov 6 A apa xou<; A, T ^.expeT 7tp«xou<; ovxac; 7tp6<; dXXfiXouc oitep eaxiv dSuvaxov. oux apa xou; A, B, T Buvaxov eaxi xexapxov dvdXoyov TtpoaeupeTv. AXXd §r] TidXiv eaxwaav oi A, B, T e^fj<; dvdXoyov, oi 8s A, r (J.r) eaxoaav npwxoi 7tp6<; dXXf|Xou<;. Xeyw, oxi 8uvax6v eaxiv auxolc; xexapxov dvdXoyov KpoaeupeTv. 6 yap B xov T 7toXXaTtXaaidaa<; xov A Ttoieixw 6 A apa xov A fjxoi [icxpel rj ou (jiexpel. (lexpeixM auxov Ttpoxepov xaxd xov E- 6 A apa xov E TioXXarcXaaidaai; xov A TteTtoirjxev. dXXd [i/]v xal 6 B xov r TioXXaTiXaaidaac; xov A 7i£Tio(r]xev 6 apa ex xwv A, E Iooq eaxi xw ex xwv B, T. dvdXoyov apa [eaxiv] «<; 6 A 7ipo<; xov B, 6 T 7tp6<; xov E- xou; A, B, T apa xexapxoc; dvdXoyov Ttpoa/]upr)xai 6 E. AXXd 8r) \ir\ (jiexpeixo 6 A xov A- Xeyw, oxi dSuvaxov eaxi xoT<; A, B, T xexapxov dvdXoyov upoaeupelv dpiiJ^ov. ei yap 8uvax6v, TipooeuprjO'dcL) 6 E - 6 apa ex xwv A, E Xaoc, eaxl xo ex xwv B, T. dXXd 6 ex xwv B, T eaxiv 6 A' xal 6 ex xfiiv A, E apa iaoc, eaxl to A. 6 A apa xov E uoXXa- TiXaaidaac; xov A 7t£TioiT]xev' 6 A apa xov A jiexpeT xaxd xov E' waxe ^expeT 6 A xov A. dXXd xal ou jiexpel" ouep axoKov. oux apa 8uvdxov eaxi xou; A, B, T xexapxov dvdXoyov n:po- aeupeTv dpi , d[i6v, oxav 6 A xov A [irj (iexpfj. dXXd 8r] oi A, B, T \±r\T£ kL,ff, eaxwaav dvdXoyov \xr\Te oi axpoi upOxoi npbz dXXf]Xou<;. xal 6 B xov T noXXaTiXaaidaa^ xov A uoielxw. b\Loitdz 8r] 8ei)(i5f]aexai, oxi ei ^ev (jiexpel 6 A xov A, 8u- vaxov eaxiv auxolc; dvdXoyov npoaeupeTv, ei 8e ou jiexpel, d8uvaxov ouep e8ei SeT^ai. as A is to B, (so) C (is) to D, and as B (is) to C, (so) D (is) to E, thus, via equality, as A (is) to C, (so) C (is) to [Prop. 7.14]. And A and C (are) prime (to one another). And (numbers) prime (to one another are) also the least (numbers having the same ratio as them) [Prop. 7.21]. And the least (numbers) measure those numbers having the same ratio as them (the same number of times), the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, A measures C, (as) the leading (measuring) the leading. And it also measures itself. Thus, A measures A and C, which are prime to one another. The very thing is impossible. Thus, it is not possible to find a fourth (number) proportional to A, B, C. And so let A, B, C again be continuously propor- tional, and let A and C not be prime to one another. I say that it is possible to find a fourth (number) propor- tional to them. For let B make D (by) multiplying C. Thus, A either measures or does not measure D. Let it, first of all, measure (D) according to E. Thus, A has made D (by) multiplying E. But, in fact, B has also made D (by) multiplying C. Thus, the (number created) from (multiplying) A, E is equal to the (number created) from (multiplying) B, C. Thus, proportionally, as A [is] to B, (so) C (is) to E [Prop. 7.19]. Thus, a fourth (number) proportional to A, B, C has been found, (namely) E. And so let A not measure D. I say that it is impossible to find a fourth number proportional to A, B, C. For, if possible, let it have been found, (and let it be) E. Thus, the (number created) from (multiplying) A, E is equal to the (number created) from (multiplying) B, C. But, the (number created) from (multiplying) B, C is D. And thus the (number created) from (multiplying) A, E is equal to D. Thus, A has made D (by) multiplying E. Thus, A measures D according to E. Hence, A measures D. But, it also does not measure (D). The very thing (is) absurd. Thus, it is not possible to find a fourth number propor- tional to A, B, C when A does not measure D. And so (let) A, B, C (be) neither continuously proportional, nor (let) the outermost of them (be) prime to one another. And let B make D (by) multiplying C. So, similarly, it can be show that if A measures D then it is possible to find a fourth (number) proportional to (A, B, C), and impossible if (A) does not measure (£>). (Which is) the very thing it was required to show. t The proof of this proposition is incorrect. There are, in fact, only two cases. Either A, B, C are continuously proportional, with A and C prime to one another, or not. In the first case, it is impossible to find a fourth proportional number. In the second case, it is possible to find a fourth proportional number provided that A measures B times C. Of the four cases considered by Euclid, the proof given in the second case is incorrect, since it only demonstrates that if A : B :: C : D then a number E cannot be found such that B : C :: D : E. The proofs given in the other three 270 ETOIXEIfiN f)'. ELEMENTS BOOK 9 cases are correct. X . Oi Tipwxoi dpidjjioi nXeiouc, eiai Travxog tou rcpoxcdevxoc; TtXr^ouc; TtptoTGJv dpidjjicov. A i 1 Hi 1 B i 1 r i 1 E A 2 i 1 — i "Eaxwaav oi TtpoxeiDevxef; Tipwxoi dpi/d^ol oi A, B, E Xeyco, oxi x65v A, B, T TtXeiouc; eiai upaixoi dpi'djioi. EiXyjcp'dw yap 6 utto xcov A, B, T eXd)(iaxoc; ^texpoujievog xdi eaxco AE, xal upoaxeia^Gj xcp AE ^ovag f) AZ. 6 5r| EZ fjxoi Tipfixog eaxiv fj ou. eaxw Ttpoxepov Ttpcoxoc;- eupr][J.evoi dpa eial rcpfiixoi dpnJ^iol oi A, B, T, EZ TiXeiouc; xfiiv A, B, r. AXXd 8/) jirj eaxw 6 EZ upaixog' utto Ttpcjxou dpa xivoc dpiiL>|jioO [icxpelxai. [lexpeiadio utio Tipwxou xou H- Xeyw, 6xi 6 H ouSevl xGv A, B, T eaxiv 6 auxoc;. ei yap Suvaxov, eaxco. oi Be A, B, T xov AE jiexpouaiv xal 6 H dpa xov AE piexprjoei. ^texpeT Se xal xov EZ- xal Xomf]v xr|v AZ ^.ovdBa jiexprjoei 6 H dpid^ioc; cov ojtep axoKov. oux dpa 6 H evi xGv A, B, T eaxiv 6 auxo<;. xal UTtoxeixai 7tp«xo<;. eupr)[ievoi dpa eiai upwxoi dpid^ol nkeiouz xoO Ttpoxe , devxo<; TtXTydou? xGv A, B, T oi A, B, T, H- onep eBei SeT^ai. Proposition 20 The (set of all) prime numbers is more numerous than any assigned multitude of prime numbers. A i 1 G' 1 B 1 C' 1 E D F i 1 1 Let A, B, C be the assigned prime numbers. I say that the (set of all) primes numbers is more numerous than A, B, C. For let the least number measured by A, B, C have been taken, and let it be DE [Prop. 7.36]. And let the unit DF have been added to DE. So EF is either prime, or not. Let it, first of all, be prime. Thus, the (set of) prime numbers A, B, C, EF, (which is) more numerous than A, B, C, has been found. And so let EF not be prime. Thus, it is measured by some prime number [Prop. 7.31]. Let it be measured by the prime (number) G. I say that G is not the same as any of A, B, C. For, if possible, let it be (the same). And A, B, C (all) measure DE. Thus, G will also measure DE. And it also measures EF. (So) G will also mea- sure the remainder, unit DF, (despite) being a number [Prop. 7.28]. The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime. Thus, the (set of) prime numbers A, B, C, G, (which is) more numerous than the assigned multitude (of prime numbers), A, B, C, has been found. (Which is) the very thing it was required to show. xa'. 'Edv dpxioi dpidjioi ottoooiouv auvxe'dwaiv, 6 oXoc; dpxioc; eaxiv. A B T A E i 1 1 1 1 Euyxeicxdwaav yap dpxioi dpid^tol OKoaoiouv oi AB, BT, TA, AE' Xeyco, Sxi oXo<; 6 AE dpxioc; eaxiv. 'Euel yap exaaxoc xcov AB, BT, TA, AE dpxioc; eaxiv, exei ^epo<; f^tair toaxe xal oXoc 6 AE e/ei [lepoc, f^iou. dpxioc; 8e dpid^toc; eaxiv 6 8[)(a Biaipou^tevoc;- dpxioc dpa eaxiv 6 AE- ouep eSei SeT^ai. Proposition 21 If any multitude whatsoever of even numbers is added together then the whole is even. A B C D E i 1 1 1 1 For let any multitude whatsoever of even numbers, AB, BC, CD, DE, lie together. I say that the whole, AE, is even. For since everyone of AB, BC, CD, DE is even, it has a half part [Def. 7.6]. And hence the whole AE has a half part. And an even number is one (which can be) divided in half [Def. 7.6]. Thus, AE is even. (Which is) 271 ETOIXEIfiN fl'. ELEMENTS BOOK 9 x(3'. 'Eav Tcepiaaol apid^oi oTcoaoioOv auvxcdwaiv, to 8e TiXfj'dot; auxov apxiov fj, 6 oXoc; dpxioc; eaxai. A B T A E i 1 1 1 1 Suyxeia'dcooav yap Tcepiaaol apid^ol 6aoi8r)icoxo0v dpxioi to TcXfjdoc: ol AB, Br, EA, AE- Xeyw, on oXoc; 6 AE dpxioc; eaxiv. 'EtccI yap exaaxoc; xwv AB, Br, TA, AE Tcepixxoc; eaxiv, dcpaipcMarjc; ^iovd8oc; dtp' exdaxou exaaxoc; xwv Xoitcov dpxioc; eaxar Saxe xal 6 auyxei^ievoc; e<; auxwv dpxioc; eaxai. eaxi 8e xal xo TtXfj'doc; xwv ^.ovdBwv apxiov. xal oXoc; dpa 6 AE dpxioc; eaxiv oTcep e8ei SeTc;ai. the very thing it was required to show. Proposition 22 If any multitude whatsoever of odd numbers is added together, and the multitude of them is even, then the whole will be even. A B C D E i 1 1 1 1 For let any even multitude whatsoever of odd num- bers, AB, BC, CD, DE, lie together. I say that the whole, AE, is even. For since everyone of AB, BC, CD, DE is odd then, a unit being subtracted from each, everyone of the remain- ders will be (made) even [Def. 7.7]. And hence the sum of them will be even [Prop. 9.21]. And the multitude of the units is even. Thus, the whole AE is also even [Prop. 9.21]. (Which is) the very thing it was required to show. xy'. 'Eav Tiepiaaol dpid^iol OTcoaoioOv auvxrdwaiv, xo 8s TiXfj'dot; aOxov Tcepiaaov fj, xal 6 oXoc; icepiaaoc; eaxai. A B T E A i 1 — i 1 — i Xuyxeia'dwaav yap OTcoaoioOv Tiepiaaol dpid^ioi, Sv xo TcXrji9oc; Tcepiaaov eaxw, ol AB, Br, EA- Xeyw, oxi xal oXoc; 6 AA icepiaaoc; eaxiv. Acpr]pr ! |a'do dico xoO IA ^tovdc; f\ AE- Xoitcoc; dpa 6 TE dpxioc; eaxiv. eaxi 8e xal 6 FA dpxioc;- xal oXoc; dpa 6 AE dpxioc; eaxiv. xai eaxi (lovac; r\ AE. Tcepiaaoc; dpa eaxiv 6 AA- oTiep e8ei SeT^ai. Proposition 23 If any multitude whatsoever of odd numbers is added together, and the multitude of them is odd, then the whole will also be odd. ABC ED i 1 1 1 — i For let any multitude whatsoever of odd numbers, AB, BC, CD, lie together, and let the multitude of them be odd. I say that the whole, AD, is also odd. For let the unit DE have been subtracted from CD. The remainder CE is thus even [Def. 7.7]. And CA is also even [Prop. 9.22]. Thus, the whole AE is also even [Prop. 9.21]. And DE is a unit. Thus, AD is odd [Def. 7.7]. (Which is) the very thing it was required to show. x5'. 'Eav dTto dpxbu dpi%ou dpxioc; d(paipei5fj, 6 Xoitcoc; dpxioc; eaxai. a r b i 1 1 Atco yap dpxiou xou AB dpxioc; dcprprja'dco 6 BE Xeyw, oxi 6 Xoitcoc; 6 EA dpxioc; eaxiv. 'Etcci yap 6 AB dpxioc; eaxiv, e^ei ^tepoc; y^iau. 8ia ^d auxd Br| xal 6 BT exei ^epoc; r^iatr Saxe xal Xoitcoc; [6 FA exei [iepoc; r^iau] dpxioc; [dpa] eaxiv 6 AT- oTiep e8ei 8eTi;ai. Proposition 24 If an even (number) is subtracted from an(other) even number then the remainder will be even. A C B For let the even (number) BC have been subtracted from the even number AB. I say that the remainder CA is even. For since AB is even, it has a half part [Def. 7.6]. So, for the same (reasons), BC also has a half part. And hence the remainder [CA has a half part]. [Thus,] AC is even. (Which is) the very thing it was required to show. 272 ETOIXEIfiN fl'. ELEMENTS BOOK 9 xe'. 'Edv duo dpxiou dpidfiou Ttepiaaoc; dcpaipe'drj, 6 Xoitcoc; Ttepiaaoc; eaxai. A r A B I 1 1 1 Atco yap dpxiou xou AB Ttepiaaoc; dcpr)pf|a , d« 6 BE Xeyco, oxi 6 Xomoc; 6 EA Ttepiaaoc; eaxiv. Acpr)pr]a'do jap and xou Br \lovolz f] EA- 6 AB dpa apxioc eaxiv. eaxi 8e xal 6 AB apxioc xal XoiTtoc; dpa 6 AA apxioc eaxiv. xai eaxi (iovdc f] EA- 6 EA dpa Ttepiaaoc eaxiv oitep e8ei Belial. Xf'. 'Eav duo Ttepiaaou dpid^iou Ttepiaaoc dcpaipe'drj, 6 XoiTtoc apxioc eaxai. A r A B I 1 1 1 Atco yap Ttepiaaou xou AB Ttepiaaoc dcpr)pr]a , do 6 BE Xey«, oxi 6 Xomoc 6 EA apxioc eaxiv. Tkei yap 6 AB Ttepiaaoc eaxiv, dcpr)pf|a , dco [lovac, f] BA- XoiTtoc dpa 6 AA apxioc eaxiv. Bid xd auxd 5r] xal 6 EA apxioc eaxiv Saxe xai XoiTtoc 6 EA apxioc eaxiv oTtep e8ei SeT^ai. 'Edv aTto Ttepiaaou dpi'd^iou apxioc dcpaipeiJrj, 6 XoiTtoc Ttepiaaoc eaxai. A A r B I 1 1 1 Atco yap Ttepiaaou xou AB apxioc dcpr]pr]a , d« 6 BE Xeyo, oxi 6 Xomoc 6 EA Ttepiaaoc eaxiv. Acpr)pr]a , dw [yap] [iovdc f) AA- 6 AB dpa apxioc; eaxiv. eaxi 8e xal 6 Br apxioc;- xal Xomoc dpa 6 EA apxioc; eaxiv. Ttepiaaoc dpa 6 EA- OTtep e8ei 5eTc;ai. XT]'. 'Eav Ttepiaaoc dpid^toc apxiov TtoXXaTtXaaidaac Ttoirj xiva, 6 yevojievoc apxioc; eaxai. Proposition 25 If an odd (number) is subtracted from an even num- ber then the remainder will be odd. A CD B For let the odd (number) BC have been subtracted from the even number AB. I say that the remainder CA is odd. For let the unit CD have been subtracted from BC. DB is thus even [Def. 7.7]. And AB is also even. And thus the remainder AD is even [Prop. 9.24]. And CD is a unit. Thus, CA is odd [Def. 7.7]. (Which is) the very thing it was required to show. Proposition 26 If an odd (number) is subtracted from an odd number then the remainder will be even. A C D B For let the odd (number) BC have been subtracted from the odd (number) AB. I say that the remainder CA is even. For since AB is odd, let the unit BD have been subtracted (from it). Thus, the remainder AD is even [Def. 7.7]. So, for the same (reasons), CD is also even. And hence the remainder CA is even [Prop. 9.24]. (Which is) the very thing it was required to show. Proposition 27 If an even (number) is subtracted from an odd num- ber then the remainder will be odd. AD C B i 1 1 1 For let the even (number) BC have been subtracted from the odd (number) AB. I say that the remainder CA is odd. [For] let the unit AD have been subtracted (from AB) . DB is thus even [Def. 7.7]. And BC is also even. Thus, the remainder CD is also even [Prop. 9.24]. CA (is) thus odd [Def. 7.7]. (Which is) the very thing it was required to show. Proposition 28 If an odd number makes some (number by) multiply- ing an even (number) then the created (number) will be even. 273 ETOIXEIfiN fl'. ELEMENTS BOOK 9 A i 1 A^ B ' 1 r i 1 Hspiaooz ydp api%6<; 6 A dpxiov xov B TtoXXa- TtXaaidaac; tov T itoishw Xeyio, oxi 6 T apxioc; eaxiv. 'End yap 6 A tov B TroXXaTtXaaidaac; tov T kstioi/jxev, 6 T dpa auyxeixai ex xoaouxtov lacov ifi B, oaai rialv sv xG A (iovdSeg. xai eaxiv 6 B apxioc 6 T apa auyxeixai z\ dpxicov. edv 6e dpxioi dpidjjiol onoaoiouv auvxe-dGaiv, 6 oXoc; apxioc; eaxiv. apxiog apa eaxiv 6 T- onep eBei 5eT<;ai. X0'. 'Edv rcepiaaoc; dpi'd^oc; Ttepiaaov dpid^tov TtoXXauXaaidc;- ac, Ttoifj Tiva, 6 yev6[ievo<; Ttepiaaoc; eaxai. A' 1 B i 1 r i 1 ilepiaaoc; yap dpid^ioc; 6 A Ttepiaaov tov B uoXXa- jcXaaidaac; tov T Tioieixw Xeyco, oxi 6 T Ttepiaaoc; eaxiv. 'Ercel yap 6 A tov B TioXXajcXaaidaac; tov T TieTcoirjxev, 6 T apa auyxeixai ex xoaouxov laov xG B, oaai eialv ev iu A ^ovdBec;. xa( eotiv exdxepoc; xGv A, B Ttepiaaoc 6 T apa auyxeixai ex TtepiaaGv dpid^iGv, Gv to TiXfj-doi; Ttepiaaov eaxiv. uoxe 6 T Ttepiaaoc; eaxiv oTtep eSei Belial. X'. 'Edv Ttepiaaoc; dpid^oc; dpTiov dpnf^ov ^Expfj, xal tov r^iauv auxou jiexprjoei. A i 1 B i 1 r i 1 Ilepiaaoc; yap dprd^ioc; 6 A dpxiov xov B ^.sxpeixw Xeyio, oxi xal xov y](itauv auxou ^exprpei. 'EtccI yap 6 A xov B [icxpel, ^texpeixM auxov xaxd xov F' Xeyco, oxi 6 T oux eaxi Ttepiaaoc;. el yap Suvaxov, eaxa>. xal eTtel 6 A xov B [iexpeT xaxd xov T, 6 A apa xov T TtoXXaTtXaaidaac; xov B TteTtofyxev. 6 B apa auyxeixai ex TtepiaaGv dpn9[iGv, Gv xo TtXTj-doc; Ttepiaaov eaxiv. 6 B apa Bi 1 Ci 1 For let the odd number A make C (by) multiplying the even (number) B. I say that C is even. For since A has made C (by) multiplying B, C is thus composed out of so many (magnitudes) equal to B, as many as (there) are units in A [Def. 7.15]. And B is even. Thus, C is composed out of even (numbers) . And if any multitude whatsoever of even numbers is added together then the whole is even [Prop. 9.21]. Thus, C is even. (Which is) the very thing it was required to show. Proposition 29 If an odd number makes some (number by) multiply- ing an odd (number) then the created (number) will be odd. A' ' B' 1 C' 1 For let the odd number A make C (by) multiplying the odd (number) B. I say that C is odd. For since A has made C (by) multiplying B, C is thus composed out of so many (magnitudes) equal to B, as many as (there) are units in A [Def. 7.15]. And each of A, B is odd. Thus, C is composed out of odd (num- bers), (and) the multitude of them is odd. Hence C is odd [Prop. 9.23]. (Which is) the very thing it was required to show. Proposition 30 If an odd number measures an even number then it will also measure (one) half of it. A i 1 B' 1 Ci 1 For let the odd number A measure the even (number) B. I say that (A) will also measure (one) half of (B). For since A measures B, let it measure it according to C. I say that C is not odd. For, if possible, let it be (odd). And since A measures B according to C, A has thus made B (by) multiplying C. Thus, B is composed out of odd numbers, (and) the multitude of them is odd. B is thus 274 ETOIXEIfiN fl'. ELEMENTS BOOK 9 Ttepiaaoc eaxiv onep axorcov UTtoxeixai yap dpxioc;. oux apa 6 r Ttepiaaoc; eaxiv dpxioc; apa eaxiv 6 T. waxe 6 A xov B ^texpeT dpxidxic;. Bid 8rj xouxo xal xov fj^iauv auxou [Lejpfiaei- oiiep e8ei BeT^ai. Xa'. 'Edv icepiaaoc; dpid^toc; upoc; xiva dpid[i6v rcpaixoc; rj, xal npoQ xov 8i7iXaa(ova auxou Kpwxoc eaxai. A' 1 B ' 1 r i 1 A' ' TLzpiaaoc, yap dpiif)[i6<; 6 A Ttpoc; xiva dpid^iov xov B npoxoc; eaxw, xou 8e B SmXaoiwv eaxto 6 E Xeyoj, 0X1 ° A [xal] upoc; xov T rcpcoxoc; eaxiv. Ei yap \ir\ eiaiv [ol A, T] Ttpfixoi, jiexpfpei xic; auxoug dpi-f^oc;. jiexpeixw, xal eaxto 6 A. xal eaxiv 6 A Ttepiaaoc;- Ttepiaaoc; apa xal 6 A. xal end 6 A Ttepiaaoc; uv xov T [icxpeT, xa[ eaxiv 6 T apxioc;, xal xov rj[iiauv apa xou T (jiexpiqaei [6 A], xou 8e T fj^uau eaxiv 6 B- 6 A apa xov B ^lexpeT. ^texpeT Be xal xov A. 6 A apa xou? A, B ^texpeT Ttpwxouc; ovxac; Ttpoc; dXXrjXouc oTtep eaxlv dSuvaxov. oux apa 6 A Ttpoc; xov T TtpGxoc; oux eaxiv. oi A, T apa TtpGxoi Ttpoc; dXXr|Xou<; eiaiv OTtep eBei 8eTc;ai. X(3'. Tcov duo 8ua8oc; 8mXaaia^ou.ev«v dpi-djiwv exaaxoc; dpxidxic; dpxioc; eaxi [iovov. A' ' B i 1 r i 1 A' ' Atco yap 8uaBoc; xfjc; A 8eBiTtXaaidaiL>Maav baoihr]- noxouv dpi^ol oi B, T, A- Xeyw, oxi oi B, T, A dpxidxic; dpxioi eiai [iovov. "Oxi [lev ouv exaaxoc; [xwv B, T, A] dpxidxic; dpxioc; eaxiv, cpavepov aTto ydp Bud8oc; eaxi BiTtXaaiaai&eic;. Xeyw, oxi xal [iovov. exxeicdw ydp [iovdc;. enel ouv aTto [iovdBoc; ojioaoiouv dpid[iol e^fjg dvdXoyov eiaiv, 6 8e [iexd xrjv [iovd8a 6 A upwxoc; eaxiv, 6 [ieyiaxoc; xfiv A, B, T, A 6 odd [Prop. 9.23]. The very thing (is) absurd. For (B) was assumed (to be) even. Thus, C is not odd. Thus, C is even. Hence, A measures B an even number of times. So, on account of this, (A) will also measure (one) half of (£>). (Which is) the very thing it was required to show. Proposition 31 If an odd number is prime to some number then it will also be prime to its double. A' 1 B 1 Ci ' D' 1 For let the odd number A be prime to some number B. And let C be double B. I say that A is [also] prime to C. For if [A and C] are not prime (to one another) then some number will measure them. Let it measure (them), and let it be D. And A is odd. Thus, D (is) also odd. And since D, which is odd, measures C, and C is even, [D] will thus also measure half of C [Prop. 9.30]. And B is half of C. Thus, D measures B. And it also measures A. Thus, D measures (both) A and B, (despite) them being prime to one another. The very thing is impossible. Thus, A is not unprime to C. Thus, A and C are prime to one another. (Which is) the very thing it was required to show. Proposition 32 Each of the numbers (which is continually) doubled, (starting) from a dyad, is an even-times-even (number) only. A' ' B 1 Ci ' D' 1 For let any multitude of numbers whatsoever, B, C, D, have been (continually) doubled, (starting) from the dyad A. I say that B, C, D are even-times-even (num- bers) only. In fact, (it is) clear that each [of B, C, D] is an even-times-even (number) . For it is doubled from a dyad [Def. 7.8]. I also say that (they are even-times-even num- bers) only. For let a unit be laid down. Therefore, since 275 ETOIXEIfiN fl'. ELEMENTS BOOK 9 A On' ouSevoc; aXXou [j.exprj'driaexai napei; ^wv A, B, T. xod eaxiv exaaxoc; xwv A, B, T apxioc 6 A apa dpxidxic; apxioc; eaxi [lovov. b[io'u^z 8f) 8e(c;o[iev, oxi [xal] exdxepoc; xwv B, T dtpxidxic; apxiog eaxi [xovov onep eSei Sei<;ai. Xy'. 'Eav dpid^ioc; xov fjjiiauv e/r) nepiaaov, dpxidxic; ne- piaaoc; eaxi [iovov. A' 1 'Apvdjidc; yap 6 A xov fjuiauv 'tyzTU> nepiaaov Xeyto, oxi 6 A dpxidxic; nepiaaoc; eaxi [iovov. "Oxi [lev ouv dpxidxic; nepiaaoc; eaxiv, cpavepov 6 yap r^iauc; auxou nepioaoc &v [iexpeT aOxov dpxidxic;, Xeyco Sr|, oxi xal \iovov. eE yap eaxai 6 A xal dpxidxic; apxioc;, [ls- xp/jOiqaexai uno dpxiou xaxd apxiov dprd^tov uoxe xal 6 f\\x\.av>c, auxou jiexp/ji&iqaexai uno dpxiou dpidjiou nepiaaoc; uv onep eaxiv axonov. 6 A apa dpxidxic; nepioaoc; eaxi [iovov onep eSei 5el<;ai. X8'. 'Eav dpii9|ji6c; \JX\ie xwv and 8ud8o<; BinXaaiai^o^evwv fj, y.r]TS xov r^iauv z^T] nepiaaov, dpxidxic; xe apxioc; eaxi xal dpxidxic; nepiaaoc;. A' 1 Api-djjioc; yap 6 A [ir|xe xGv dno 8ud8oc; 8mXaaia£o[iivov eaxw [lifts xov fjuiauv tyixu nepiaaov Xeyw, oxi 6 A dpxidxic; xe eaxiv apxioc; xal dpxidxic; nepiaaoc- "Oxi [lev ouv 6 A dpxidxic; eaxiv apxioc;, cpavepov xov yap rjuiauv oux exei nepiaaov. Xey« 8rj, oxi xal dpxidxic; ne- piaaoc eaxiv. eav yap xov A xeuvwuev 8()(a xal xov fjuiauv auxou 8[)(a xal xouxo del noifiuev, xaxavxrjaouev zic, xiva dpiduov nepiaaov, oc uexpf|aei xov A xaxd apxiov dpiduov. ei yap ou, xaxavxyjaouev eic 8ud8a, xal eaxai 6 A xwv dno 8ud8oc 8mXaaia£ouev«v onep oux unoxeixai. waxe 6 A dpxidxic; nepiaaov eaxiv. e8e[)fdr] 8e xal dpxidxic; apxioc;. 6 A apa dpxidxic; xe apxioc; eaxi xal dpxidxic; nepiaaoc onep e8ei 8elc;ai. any multitude of numbers whatsoever are continuously proportional, starting from a unit, and the (number) A af- ter the unit is prime, the greatest of A, B, C, D, (namely) D, will not be measured by any other (numbers) except A, B, C [Prop. 9.13]. And each of A, B, C is even. Thus, D is an even-time-even (number) only [Def. 7.8]. So, similarly, we can show that each of B, C is [also] an even- time-even (number) only. (Which is) the very thing it was required to show. Proposition 33 If a number has an odd half then it is an even-time- odd (number) only. A i 1 For let the number A have an odd half. I say that A is an even-times-odd (number) only. In fact, (it is) clear that (A) is an even-times-odd (number). For its half, being odd, measures it an even number of times [Def. 7.9]. So I also say that (it is an even-times-odd number) only. For if A is also an even-times-even (number) then it will be measured by an even (number) according to an even number [Def. 7.8]. Hence, its half will also be measured by an even number, (despite) being odd. The very thing is absurd. Thus, A is an even-times-odd (number) only. (Which is) the very thing it was required to show. Proposition 34 If a number is neither (one) of the (numbers) doubled from a dyad, nor has an odd half, then it is (both) an even-times-even and an even-times-odd (number) . A i 1 For let the number A neither be (one) of the (num- bers) doubled from a dyad, nor let it have an odd half. I say that A is (both) an even-times-even and an even- times-odd (number). In fact, (it is) clear that A is an even-times-even (num- ber) [Def. 7.8]. For it does not have an odd half. So I say that it is also an even-times-odd (number) . For if we cut A in half, and (then cut) its half in half, and we do this continually, then we will arrive at some odd num- ber which will measure A according to an even number. For if not, we will arrive at a dyad, and A will be (one) of the (numbers) doubled from a dyad. The very oppo- site thing (was) assumed. Hence, A is an even-times-odd (number) [Def. 7.9]. And it was also shown (to be) an even-times-even (number). Thus, A is (both) an even- times-even and an even-times-odd (number) . (Which is) 276 ETOIXEIfiN fl'. ELEMENTS BOOK 9 Xe'. 'Edv fiaiv 6aoiSr)7ioxouv apidfiol kc,r\c, dvdXoyov, dcpai- pcdfiai 8e duo xe xou Seuxepou xal tou ea)(dxou laoi iu Ttpcbxw, eaxai ox f) xou 8euxepou UTiepo/r) Ttpoc; xov Ttpfixov, ouxcoe; f] tou ea/dxou (mepoxr) Ttpoc; xouc; Tipo eauxou ndvxac;. A' 1 b h r I 1 1 the very thing it was required to show. Proposition 35 f If there is any multitude whatsoever of continually proportional numbers, and (numbers) equal to the first are subtracted from (both) the second and the last, then as the excess of the second (number is) to the first, so the excess of the last will be to (the sum of) all those (num- bers) before it. A' ' B G C i 1 — i A 1 E A K Z i — i — i — i 1 'Eaxcoaav oTioaoiSrjTtoxouv dpi/d^io! dvdXoyov oi A, Br, A, EZ dcp^o^evoi aTto eXaxiaxou xou A, xdi dcpr)pr]C7d« dno xou Br xdi xou EZ xcb A loot; exdxepoc; xfiv BH, Z0- Xeyw, oxi eaxiv cbe; 6 BT npoc, xov A, ouxioc; 6 E0 Ttpoc; xouc; A, Br, A. Keicdco y&P t<P Br Xaoc, 6 ZK, xcp 8e A Xaoc, 6 ZA. xal inei 6 ZK xfi Br Xaoc, eaxiv, Sv 6 Z0 iu BH lao? eaxiv, Xoiitoc; dpa 6 0K Xouuo xQ HT eaxiv laoc;. xdi eitei eaxiv ox 6 EZ Ttpoc; xov A, ouxox 6 A Ttpoc; xov Br xdi 6 Br Ttpoc; xov A, iao<; 8e 6 ^tev A xw ZA, 6 8e Br xw ZK, 6 8e A iw ZO, eaxiv dpa ox 6 EZ Ttpoc; xov ZA, ouxox 6 AZ Ttpoc; xov ZK xdi 6 ZK Ttpoc; xov Z0. SieXovxi, ox 6 EA Ttpoc; xov AZ, ouxox 6 AK Ttpoc; xov ZK xdi 6 KG Ttpoc; xov Z0. eaxiv dpa xal ox zXc, xwv r)you[ievojv npoc eva xwv eTtojievojv, ouxox aTtavxec; oi rjyou^ievoi npoc aTtavxac; xouc; CTto|ievou<;- eaxiv dpa ox 6 KG Ttpoc; xov Z0, ouxox oi EA, AK, KG Ttpoc; xou? AZ, ZK, 6Z. iaoc 8e 6 yttv K9 xo> TH, 6 Se Z0 xo> A, oi Se AZ, ZK, 6Z xol? A, Br, A- eaxiv dpa ox 6 TH Ttpoc; xov A, ouxox 6 E0 Ttpoc; xoix A, Br, A. eaxiv dpa ox f] xou 8euxepou UTiepoxr] npoc xov Ttpwxov, ouxox f\ xou ea)(dxou UTtepoxr] Ttpoc; xoix Ttpo eauxou Ttdvxac oitep e8ei 8eT^ai. t This proposition allows us to sum a geometric series of the form a, (ar- a)/a = (ar n - a)/S„. Hence, 5„ = a (r n - l)/(r - 1). X<r'. 'Edv aTto [iovd8oc; onoaoiouv dpidjiol e^fjc; exxeiDwaiv ev xfj BiTtXaaiovi dvaXoyia, eox ou 6 au^iTtac; auvxe-deic; Ttpfiixoc; yevrjxai, xal 6 au^iTtac; era xov ea)(axov TtoXXaTtXaaiaaiJelc; D 1 E L K H F i 1 1 1 1 Let A, BC, D, EF be any multitude whatsoever of continuously proportional numbers, beginning from the least A. And let BG and FH, each equal to A, have been subtracted from BC and EF (respectively). I say that as GC is to A, so EH is to A, BC, D. For let FK be made equal to BC, and FL to D. And since FK is equal to BC, of which FH is equal to BG, the remainder 7? K is thus equal to the remainder GC. And since as EF is to £>, so D (is) to BC, and BC to A [Prop. 7.13], and D (is) equal to FL, and BC to Bif, and A to FiJ, thus as EF is to FL, so LF (is) to FJf, and FK to Fff. By separation, as EL (is) to LF, so £AT (is) to FK, and FiJ to FiJ [Props. 7.11, 7.13]. And thus as one of the leading (numbers) is to one of the following, so (the sum of) all of the leading (numbers is) to (the sum of) all of the following [Prop. 7.12]. Thus, as KH is to FH, so EL, LK, KH (are) to LF, FK, HF. And KH (is) equal to CG, and FH to A, and LF, FK, HF to D, BC, A. Thus, as CG is to A, so EH (is) to D, BC, A. Thus, as the excess of the second (number) is to the first, so the excess of the last (is) to (the sum of) all those (numbers) before it. (Which is) the very thing it was required to show. ar, ar 2 , ar 3 , ■ ■ ■ ar"^ 1 . According to Euclid, the sum S n satisfies Proposition 36 f If any multitude whatsoever of numbers is set out con- tinuously in a double proportion, (starting) from a unit, until the whole sum added together becomes prime, and 277 ETOIXEIfiN fl'. ELEMENTS BOOK 9 Ttoifj xiva, 6 yev6^ievo<; xeXeioc; eaxai. Atco yap ^iovd8o<; exxeia-dwaav 6ooi8t)tioto0v dpi-d^- ol ev xfj SmXaaiovi dvaXoyia, ewe; ou 6 aujjmac; auvxei&eic; 7tp£>xo<; yevrjxai, oi A, B, T, A, xai x£> aujiiravxi iao? eaxw 6 E, xal 6 E xov A TtoXXaTtXaaidaac; xov ZH Ttoieixco. Xeyco, oxi 6 ZH xeXeioc eaxiv. B i — i r i ' A | 1 the sum multiplied into the last (number) makes some (number), then the (number so) created will be perfect. For let any multitude of numbers, A, B, C, D, be set out (continuouly) in a double proportion, until the whole sum added together is made prime. And let E be equal to the sum. And let E make FG (by) multiplying D. I say that FG is a perfect (number) . B 1 C' ' D 1 E' — ' N K A 1 Mi 1 Z H H I 1 1 O ' n ' "Oaoi ydp eiaiv oi A, B, T, A xfi> rcXrydei, xoaouxoi duo xou E £iXf](pTf)«aav ev xfj SmXaaiovi dvaXoyia oi E, 9K, A, M - 8i' i'aou dpa laxlv &>c, 6 A Ttp6<; xov A, ouxcoc; 6 E npoc; xov M. 6 dpa ex x«v E, A i'aoc; eaxi x£> ex xwv A, M. xai eaxiv 6 ex xQv E, A 6 ZH- xai 6 ex xuv A, M dpa eaxiv 6 ZH. 6 A dpa xov M TtoXXaTtXaaidaac; xov ZH TteTtoirjxev 6 M dpa xov ZH ^.expeT xaxa xa<; ev x£> A ^.ovdBac;. xai eaxi Suae; 6 A- 8iTtXdaio<; dpa eaxiv 6 ZH xou M. eiai 8e xai oi M, A, 0K, E e^rjej SmXdoioi dXXf|X«v oi E, 0K, A, M, ZH dpa e^r\Q dvdXoyov eiaiv ev xfj BiTtXaaiovi dvaXoyia. dcpr)pf|a , d« 8rj duo xou Beuxepou xou 6K xai xou eo^axou xou ZH xu Tipcoxw xw E i'aoc; exdxepo<; xwv 9N, ZS - eaxiv dpa d>c f\ xou Seuxepou dprd|iou UKepo^f] 7ip6<; xov Tipfixov, ouxoc; f) xou ea)(dxou UTiepo)(r) 7ipo<; xouc; Ttpo eauxou Ttdvxac;. eaxiv dpa &><; 6 NK 7ip6<; xov E, ouxwc; 6 SH 7ip6<; xouc; M, A, KG, E. xai eaxiv 6 NK i'aoc; x£> E - xai 6 SH dpa Taoc; eaxi xdic M, A, 8K, E. eaxi 8e xai 6 ZS xw E iaoz, 6 8e E xou; A, B, r, A xai xfj ^tovd8i. oXoc; dpa 6 ZH iaoc; eaxi xou; xe E, 9K, A, M xai xou; A, B, T, A xai xfj ^iovd8i- xai ^lexpeTxai bn auxwv. Xeyw, oxi xai 6 ZH viz' ou8evo<; dXXou ^exprj-driaexai nape? xwv A, B, T, A, E, 9K, A, M xai xfjc ^.ovdBoc;. ei yap 8uvax6v, ^.expeixw xic; xov ZH 6 O, xai 6 O [LrpeVi xQv A, B, T, A, E, 9K, A, M eaxw 6 auxocj. xai oadxu; 6 O xov ZH jiexpel, xoaauxai ^iovd8e<; Ei ' H N K I 1 1 L i 1 Mi ' F O G P i 1 Q For as many as is the multitude of A, B, C, D, let so many (numbers), E, HK, L, M, have been taken in a double proportion, (starting) from E. Thus, via equal- ity, as A is to D, so E (is) to M [Prop. 7.14]. Thus, the (number created) from (multiplying) E, D is equal to the (number created) from (multiplying) A, M. And FG is the (number created) from (multiplying) E, D. Thus, FG is also the (number created) from (multiplying) A, M [Prop. 7.19]. Thus, A has made FG (by) multiplying M. Thus, M measures FG according to the units in A. And A is a dyad. Thus, FG is double M. And M, L, HK, E are also continuously double one another. Thus, E, HK, L, M, FG are continuously proportional in a double proportion. So let HN and FO, each equal to the first (number) E, have been subtracted from the second (number) HK and the last FG (respectively). Thus, as the excess of the second number is to the first, so the ex- cess of the last (is) to (the sum of) all those (numbers) before it [Prop. 9.35]. Thus, as NK is to E, so OG (is) to M, L, KH, E. And NK is equal to E. And thus OG is equal to M, L, HK, E. And FO is also equal to E, and E to A, B, C, D, and a unit. Thus, the whole of FG is equal to E, HK, L, M, and A, B, C, D, and a unit. And it is measured by them. I also say that FG will be 278 ETOIXEIfiN fl'. ELEMENTS BOOK 9 eaxtoaav ev iu II ■ 6 II apa xov O rcoXXaTiXaaidaat; xov ZH Tten:o[r)xev. dXXd [jnrjv xai 6 E xov A TtoXXaTtXaaidaac; xov ZH 7CE7ioir]xev eaxiv apa cbc; 6 E Tipoc xov II, 6 O Tipoc; xov A. xal etcel duo ^lovdBoc; e<;rj<; dvdXoyov eiaiv ol A, B, T, A, 6 A apa On' ouSevog aXXou dpi-djioO jj.sxprj'diqaexai nape? xcov A, B, T. xal UTioxeixai 6 O ouSevi iSv A, B, T 6 auxoc oux apa ^.expf|aei 6 O xov A. dXX' «<; 6 O 7tp6<; xov A, 6 E 7ip6<; xov II- ou8e 6 E apa xov II ^.expel. xa[ eaxiv 6 E 7ipoxo<;- n&z 8e Ttpwxoc; dpi-djaot; npbc, anavxa, 6v \ir\ jiexpe'i, 7tp£>x6<; [eaxiv]. oi E, II apa npGxoi Kpog dXXr|Xou<; eiaiv. oi 8e npwxoi xal eXd)(iaxoi, oi 8e eXd)(iaxoi ^.expouai xou<; xov auxov Xoyov ey^ovxac; iadxic; o xe rjyounevoi; xov rjyou^ievov xal 6 £iz6[ievoci xov CTtojievov xa[ eaxiv «<; 6 E Ttp6<; xov n, 6 O Tip6<; xov A. iadxi<; apa 6 E xov O (jiexpeT xal 6 II xov A. 6 8e A Cm' ouBevoc dXXou ^expeTxai nape? xov A, B, T- 6 n apa evl xov A, B, T eaxiv 6 auxoc;. eaxco twBo auxoc;. xal oaoi eiaiv oi B, T, A xw TtXrydei xoaoOxoi eiXTjcpiJoaav aTio xou E oi E, 6K, A. xai eiaiv oi E, 9K, A toXq B, T, A ev xw auxw Xoyw 8i' taou apa eaxiv w?6B Kpoc; xov A, 6 E 7ip6<; xov A. 6 apa ex xwv B, A Taoc eaxl xo ex xQv A, E- dXX' 6 ex xwv A, E iao<; eaxi x« ex xwv II, O- xal 6 ex xwv n, O apa Xaoc, eaxl iw ex xwv B, A. eaxiv apa w<; 6 II 7ip6<; xov B, 6 A 7ip6<; xov O. xai eaxiv 6 II xw B 6 auxoc - xal 6 A apa xw O eaxiv 6 auxoc ojiep d86vaxov 6 yap O UTioxeixai jir]8evl xwv exxeijievcov 6 auxoc oux apa xov ZH \LZxpr\oz\. tiz dpn!)^6<; Kape? xCSv A, B, T, A, E, 9K, A, M xal xfj? jiovd8o<;. xal eBeix/] 6 ZH xoI<; A, B, T, A, E, 0K, A, M xal xfj jiovdBi Xaoc,. xeXeioi; 8e dpiif)[i6<; eaxiv 6 xou; eauxou [lepeaiv Xaoz wv xeXeioc apa eaxiv 6 ZH- oTiep e8ei 8el^ai- measured by no other (numbers) except A, B, C, D, E, HK, L, M, and a unit. For, if possible, let some (num- ber) P measure FG, and let P not be the same as any of A, B, C, D, E, HK, L, M. And as many times as P measures FG, so many units let there be in Q. Thus, Q has made FG (by) multiplying P. But, in fact, E has also made FG (by) multiplying D. Thus, as E is to Q, so P (is) to D [Prop. 7.19]. And since A, B, C, D are con- tinually proportional, (starting) from a unit, D will thus not be measured by any other numbers except A, B, C [Prop. 9.13]. And P was assumed not (to be) the same as any of A, B, C. Thus, P does not measure D. But, as P (is) to D, so E (is) to Q. Thus, E does not mea- sure Q either [Def. 7.20]. And E is a prime (number). And every prime number [is] prime to every (number) which it does not measure [Prop. 7.29]. Thus, E and Q are prime to one another. And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (num- bers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. And as E is to Q, (so) P (is) to D. Thus, E measures P the same number of times as Q (measures) D. And D is not measured by any other (numbers) except A, B, C. Thus, Q is the same as one of A, B, C. Let it be the same as B. And as many as is the multitude of B, C, D, let so many (of the set out numbers) have been taken, (start- ing) from E, (namely) E, HK, L. And E, HK, L are in the same ratio as B, C, D. Thus, via equality, as B (is) to D, (so) E (is) to L [Prop. 7.14]. Thus, the (number created) from (multiplying) B, L is equal to the (num- ber created) from multiplying D, E [Prop. 7.19]. But, the (number created) from (multiplying) D, E is equal to the (number created) from (multiplying) Q, P. Thus, the (number created) from (multiplying) Q, P is equal to the (number created) from (multiplying) B, L. Thus, as Q is to B, (so) L (is) to P [Prop. 7.19]. And Q is the same as B. Thus, L is also the same as P. The very thing (is) impossible. For P was assumed not (to be) the same as any of the (numbers) set out. Thus, FG cannot be measured by any number except A, B, C, D, E, HK, L, M, and a unit. And FG was shown (to be) equal to (the sum of) A, B, C, D, E, HK, L, M, and a unit. And a perfect number is one which is equal to (the sum of) its own parts [Def. 7.22]. Thus, FG is a perfect (number). (Which is) the very thing it was required to show. t This proposition demonstrates that perfect numbers take the form 2 n 1 (2 n — 1) provided that 2 n — 1 is a prime number. The ancient Greeks knew of four perfect numbers: 6, 28, 496, and 8128, which correspond to n = 2, 3, 5, and 7, respectively. 279 280 ELEMENTS BOOK 10 Incommensurable Magnitudes^ tThe theory of incommensurable magntidues set out in this book is generally attributed to Theaetetus of Athens. In the footnotes throughout this book, k, k' , etc. stand for distinct ratios of positive integers. 281 ETOIXEIftN i'. ELEMENTS BOOK 10 "Opoi. a'. Eujijiexpa jieye-dT] Xeyexai ™ T <?> auxG [xexpco fis- xpoujieva, dau^expa 56, 5v [ir)8ev evSexsxai xoivov [isxpov yevecrdai. P'. EMeTai SuvdjieL au^expoi ricnv, oxav xd an auxfiiv xexpdywva xw ocuxw X W P'<P ^ETpfjxai, dau^expoi 8e, oxav xou; dm' auxwv xexpaycovou; p]5ev £v8e)(r)xai x w p' ov xoivov ^.fxpov yevecrdoa. y'. Touxwv Otioxei^ievwv Seixvuxai, oxi xrj Ttpoxrfteiar] su-Ma uudpxouaiv £015610(1 TtXyydei ameipoi au^iexpoi xe xod dau^expoi ai [ie\> [i^xei ^tovov, ai 8s xal Suvd^iei. xaXdcrdM ouv f] [iev Tipoxe-delaa eu-fMa prjxr], xal ai xauxr] au^expoi eixs [xr]X£i xal Suvd^iei eixe 5uvd^i£i \i6mov pr]xa(, ai 5e xauxr) dau[i^t£xpoi dXoyoi xaXdcrdcoaav. 8'. Kai xo [Lev duo xrjc; Ttpoxe'fkiaric; EU'ddat; xexpdyo- vov prjxov, xal xd xouxw au^erxpa p/]xd, xd 5e xouxw dau[i^i£xpa aXoya xaXeio'dw, xal ai 5uvd^evai auxd dXoyoi, el [lev xexpdywva d'r), auxal ai TtXerupai, el 8e exepd xiva eu-duypa^ia, ai laa auxoTc xexpdywva dvaypdcpouaai. Definitions 1. Those magnitudes measured by the same measure are said (to be) commensurable, but (those) of which no (magnitude) admits to be a common measure (are said to be) incommensurable. t 2. (Two) straight-lines are commensurable in square* when the squares on them are measured by the same area, but (are) incommensurable (in square) when no area admits to be a common measure of the squares on them. § 3. These things being assumed, it is proved that there exist an infinite multitude of straight-lines commensu- rable and incommensurable with an assigned straight- line — those (incommensurable) in length only, and those also (commensurable or incommensurable) in squared Therefore, let the assigned straight-line be called ratio- nal. And (let) the (straight-lines) commensurable with it, either in length and square, or in square only, (also be called) rational. But let the (straight-lines) incommensu- rable with it be called irrational.* 4. And let the square on the assigned straight-line be called rational. And (let areas) commensurable with it (also be called) rational. But (let areas) incommensu- rable with it (be called) irrational, and (let) their square- roots* (also be called) irrational — the sides themselves, if the (areas) are squares, and the (straight-lines) describ- ing squares equal to them, if the (areas) are some other rectilinear (figure) J t In other words, two magnitudes a and l3 are commensurable if a : (3 :: 1 : k, and incommensurable otherwise, t Literally, "in power". § In other words, two straight-lines of length a and are commensurable in square if a : (3 :: 1 : A; 1 / 2 , and incommensurable in square otherwise. Likewise, the straight-lines are commensurable in length if a : /3 :: 1 : k, and incommensurable in length otherwise. 11 To be more exact, straight-lines can either be commensurable in square only, incommensurable in length only, or commenusrable/incommensurable in both length and square, with an assigned straight-line. * Let the length of the assigned straight-line be unity. Then rational straight-lines have lengths expressible as k or A; 1 / 2 , depending on whether the lengths are commensurable in length, or in square only, respectively, with unity. All other straight-lines are irrational. $ The square-root of an area is the length of the side of an equal area square. II The area of the square on the assigned straight-line is unity. Rational areas are expressible as A. All other areas are irrational. Thus, squares whose sides are of rational length have rational areas, and vice versa. a. . Auo [iEyE'dcSv dviawv exxei|jiev«v, edv duo xou y.eiZ,ovoz dcpaipsiS/j \±eXt,o\ f\ xo rj^iau xal xou xaxaXsurojievou \±elZ,o\i fj xo fjjjiiau, xal xouxo del yiyvrjxai, XeicpiD^asxal xi \ieye , doc;, o eaxai eXaoaov xou exxa^ievou eXdaaovo<; \ieye-Qouc,. "Eax« 8uo [leye'dr] aviaa xd AB, T, Sv [Le%ov xo AB- Proposition V If, from the greater of two unequal magnitudes (which are) laid out, (a part) greater than half is sub- tracted, and (if from) the remainder (a part) greater than half (is subtracted), and (if) this happens continually, then some magnitude will (eventually) be left which will 282 ETOIXEIftN i'. ELEMENTS BOOK 10 Xeyw, on, eav duo xou AB dcpaipcdfj [icTi^ov fj to fjjiiau xdi xoO xaxaXemo^tevou ueTCov fj to fj^iiau, xdi xouxo del y LYVY]Tai, XeicpiJfiaexai tl ^eyei!}oc;, o eaxai eXaaaov xou T ^eyei&ouc;. AK B be less than the lesser laid out magnitude. Let AB and C be two unequal magnitudes, of which (let) AB (be) the greater. I say that if (a part) greater than half is subtracted from AB, and (if a part) greater than half (is subtracted) from the remainder, and (if) this happens continually, then some magnitude will (eventu- ally) be left which will be less than the magnitude C. A K H B r' ' i 1 1 1 A Z H E To T ydp TioXXaTtXaaiaCo^evov eaxai Ttoxe xou AB (jiEiCov. TieTioXXomXaaidcTdw, xdi eaxw xo AE xou usv T TtoXXarcXaaiov, xou Be AB jieli^ov, xdi Sirjpfja'dco xo AE eic, xd ifi T iaa xd AZ, ZH, HE, xdi dcprjprjadco drco \xe\ xou AB [isi^ov fj xo f)(iiou xo BO, duo Be xou AO jjieT^ov fj xo fj[iiau xo OK, xdi xouxo del yiyveadio, ewe; dv di ev xai AB Biaipeaeu; loo7xX/)i9eTc yevcovxai xau; ev xcp AE Biaipeaeaiv. "Eaxwaav ouv di AK, KO, OB Biaipeaeu; laonXiQ'&eTe; ouaai xau; AZ, ZH, HE - xdi cuel (lel^ov eaxi xo AE xou AB, xdi dcpf]pr]xai dno jiev xou AE eXaaaov xou f](iiaeco<; xo EH, aTio Be xou AB [lei^ov fj xo fj^iiau xo BO, Xoitiov dpa xo HA Xoitcou xou OA jieT^ov eaxiv. xdi enel jiel^ov eaxi xo HA xou OA, xdi dcpf)pr]xai xou [iev HA fj|jiiau xo HZ, xou Be OA (icT^ov fj xo fjjiiau xo OK, Xomov dpa xo AZ Xoittou xou AK jiel^ov eaxiv. Taov Be xo AZ x£i r - xdi xo T dpa xou AK [icT^ov eoxiv. eXaaaov dpa xo AK xou T. KaxaXeinexai dpa duo xou AB jieye-douc; xo AK (leyeOoc; eXaaaov dv xou exxeijjievou eXdaaovoc (icye'dou^ xou E onep eBei BeT^ai. — o^ioimc; Be Beix^fpexai, xav f][iiar] fj xd dcpaipou^ieva. D F G E For C, when multiplied (by some number), will some- times be greater than AB [Def. 5.4]. Let it have been (so) multiplied. And let DE be (both) a multiple of C, and greater than AB. And let DE have been divided into the (divisions) DF, FG, GE, equal to C. And let BH, (which is) greater than half, have been subtracted from AB. And (let) HK, (which is) greater than half, (have been subtracted) from AH. And let this happen continu- ally, until the divisions in AB become equal in number to the divisions in DE. Therefore, let the divisions (in AB) be AK, KH, HB, being equal in number to DF, FG, GE. And since DE is greater than AB, and EG, (which is) less than half, has been subtracted from DE, and BH, (which is) greater than half, from AB, the remainder GD is thus greater than the remainder HA. And since GD is greater than HA, and the half GF has been subtracted from GD, and HK, (which is) greater than half, from HA, the remain- der DF is thus greater than the remainder AK. And DF (is) equal to C. C is thus also greater than AK. Thus, AK (is) less than C. Thus, the magnitude AK, which is less than the lesser laid out magnitude C, is left over from the magnitude AB. (Which is) the very thing it was required to show. — (The theorem) can similarly be proved even if the (parts) subtracted are halves. t This theorem is the basis of the so-called method of exhaustion, and is generally attributed to Eudoxus of Cnidus. P'- 'Edv Buo ^eyei9Gv [exxeijieviov] dviawv dvducpaipou^ievou del xou eXdaaovot; duo xou ^.ei^ovoc; xo xaxaXeiTio^ievov ^iT)8e7ioxe xaxa^texpfj xo npo eauxou, dau^expa eaxai xd ^eyc'dr). Auo yap jieye-daiv ovxtov dviawv xwv AB, TA xdi eXdaaovoc; xou AB dvducpaipou^ievou del xou eXdaaovoc; duo xou [iei^ovoc; xo nepiXemojievov [irjBejraxe xaxa^e- Proposition 2 If the remainder of two unequal magnitudes (which are) [laid out] never measures the (magnitude) before it, (when) the lesser (magnitude is) continually subtracted in turn from the greater, then the (original) magnitudes will be incommensurable. For, AB and CD being two unequal magnitudes, and AB (being) the lesser, let the remainder never measure 283 ETOIXEIftN i'. ELEMENTS BOOK 10 xpeixo to Tipo eauxou- Xeyw, oxi dauuuexpd eaxi xd AB, TA ueye'dr). AH B i — i 1 1 E' ' i 1 1 1 r z a El ydp laxi auuuexpa, uexprpei xi auxa ueycdo<;. ue- xpeixw, ei 8uvax6v, xod eaxw xo E- xal xo uev AB xo ZA xaxauexpouv XeiTiexo eauxou eXaaaov xo TZ, xo 8e TZ xo BH xaxauexpouv Xeircexo eauxou eXaaaov xo AH, xal xouxo del yivecrdco, ecog ou Xeicp^fj xi ueycdoc;, 6 eaxiv eXaaaov xou E. yeyovexco, xal XsXsicp'Ow xo AH eXaaaov xou E. insi ouv xo E xo AB uexpeT, dXXd xo AB xo AZ uexpel, xai xo E apa xo ZA uexpr|aei. uexpeT 8e xal 6Xov xo TA- xal Xomov apa xo rZ uexpiqaei. dXXd xo TZ xo BH uexpeT- xal xo E apa xo BH uexpeT. uexpeT 8e xal oXov xo AB- xal Xomov apa xo AH uexprpei, xo ueT^ov xo eXaaaov onep eaxlv dSuvaxov. oux apa xd AB, TA ueye'dr] uexprpei xi ueyeiSoc;- dauuuexpa apa eaxl xd AB, FA ueyeiSr). 'Edv apa 8uo ueyei!)Gv dviaiov, xal xd e^fjg. t The fact that this will eventually occur is guaranteed by Prop. 10.1. the (magnitude) before it, (when) the lesser (magnitude is) continually subtracted in turn from the greater. I say that the magnitudes AB and CD are incommensurable. A G B i 1 1 1 Ei 1 C F D For if they are commensurable then some magnitude will measure them (both). If possible, let it (so) measure (them), and let it be E. And let AB leave CF less than itself (in) measuring FD, and let CF leave AG less than itself (in) measuring BG, and let this happen continually, until some magnitude which is less than E is left. Let (this) have occurred, ^ and let AG, (which is) less than E, have been left. Therefore, since E measures AB, but AB measures DF, E will thus also measure FD. And it also measures the whole (of) CD. Thus, it will also mea- sure the remainder CF. But, CF measures BG. Thus, E also measures BG. And it also measures the whole (of) AB. Thus, it will also measure the remainder AG, the greater (measuring) the lesser. The very thing is impos- sible. Thus, some magnitude cannot measure (both) the magnitudes AB and CD. Thus, the magnitudes AB and CD are incommensurable [Def. 10.1]. Thus, if ... of two unequal magnitudes, and so on y'. Proposition 3 Auo ueycdcov auuuexpcov Boftevxcov xo ueyiaxov auxfiv To find the greatest common measure of two given xoivov uexpov eupeTv. commensurable magnitudes. AZ B A F B i — i — i — i — i i— i — i — i — i r E ACE D Hi ' 'Eaxw xd 5oi9evxa 8uo ueye-dr) auuuexpa xd AB, TA, Sv eXaaaov xo AB- Bel 8f) xov AB, TA xo ueyiaxov xoivov uexpov eupeTv. To AB ydp uiycdoc; rjxoi uexpeT xo TA fj ou. et uev ouv uexpel, uexpel 8e xal eauxo, xo AB apa xfiv AB, TA xoivov uexpov eaxiv xal cpavepov, oxi xal ueyiaxov. ueTCov ydp xou AB ueyei!)ou<; xo AB ou uexprpei. Mr) uexpeixw 5r] xo AB xo FA. xal dvducpaipouuevou del xou eXdaaovoc; duo xou ueiCovoc;, xo TtepiXeiitouevov uexprpei Ttoxe xo Tipo eauxou 8ia xo \ir) elvai dauuuexpa xd AB, TA- xal xo uev AB xo EA xaxauexpouv Xcikcxw eauxou Gi ' Let AB and CD be the two given magnitudes, of which (let) AB (be) the lesser. So, it is required to find the greatest common measure of AB and CD. For the magnitude AB either measures, or (does) not (measure), CD. Therefore, if it measures (CD), and (since) it also measures itself, AB is thus a common mea- sure of AB and CD. And (it is) clear that (it is) also (the) greatest. For a (magnitude) greater than magnitude AB cannot measure AB. So let AB not measure CD. And continually subtract- ing in turn the lesser (magnitude) from the greater, the 284 ETOIXEIfiN i'. ELEMENTS BOOK 10 sXaaaov to Er, to 8e Er to ZB xaxa^iexpouv Xemexo EauxoD eXaaaov xo AZ, xo 8e AZ xo EE ^texpeixw. Tkei ouv xo AZ xo EE [iexpei, dXXa xo EE xo ZB jxexpel, xal xo AZ apa xo ZB ^exprjaet. jiexpei 8e xal eauxo' xal oXov apa xo AB ^exprpei xo AZ. dXXa xo AB xo AE [lexpsi- xal xo AZ apa xo EA ^expiqoei. piexpei 8e xal xo EE- xai oXov apa xo TA [isxpel' xo AZ apa xaiv AB, EA xoivov ^.fxpov eaxiv. Xeyw 8Vj, oxi xal jieyiaxov. £ i Y<*P t 1 ^ eaxai xi [isys'doc jiei£ov xou AZ, o ^sxpfjaei xd AB, EA. eax« xo H. excel ouv xo H xo AB ^.sxpei, dXXa xo AB xo EA (lexpsi, xal xo H apa xo EA ^exprjaei. ^lExpeT Se xal oXov xo EA- xal Xomov apa xo TE [lexprpei xo H. dXXa xo TE xo ZB ^.expsT' xal xo H apa xo ZB jiexprjaei. ^.expel 8e xal oXov xo AB, xal Xomov xo AZ (jiexprpei, xo ^.el^ov xo IXaaaov oTiep eaxlv dSuvaxov. oux apa ^leT^ov xi [leye-Qoc, xou AZ xd AB, EA (jiexpr]0£i- xo AZ apa xo5v AB, EA xo jieyiaxov xoivov ^.expov scrav. Auo apa [isyeiSGv au^iexptov Bo-dcvxcov x£Sv AB, EA xo ^icyiaxov xoivov jiexpov r]upr)xai- onsp cBei Ba^ai. remaining (magnitude) will (at) some time measure the (magnitude) before it, on account of AB and CD not be- ing incommensurable [Prop. 10.2]. And let AB leave EC less than itself (in) measuring ED, and let EC leave AF less than itself (in) measuring FB, and let AF measure CE. Therefore, since AF measures CE, but CE measures FB, AF will thus also measure FB. And it also mea- sures itself. Thus, AF will also measure the whole (of) AB. But, AB measures DE. Thus, AF will also mea- sure ED. And it also measures CE. Thus, it also mea- sures the whole of CD. Thus, AF is a common measure of AB and CD. So I say that (it is) also (the) greatest (common measure) . For, if not, there will be some mag- nitude, greater than AF, which will measure (both) AB and CD. Let it be G. Therefore, since G measures AB, but AB measures ED, G will thus also measure ED. And it also measures the whole of CD. Thus, G will also mea- sure the remainder CE. But CE measures FB. Thus, G will also measure FB. And it also measures the whole (of) AB. And (so) it will measure the remainder AF, the greater (measuring) the lesser. The very thing is im- possible. Thus, some magnitude greater than AF cannot measure (both) AB and CD. Thus, AF is the greatest common measure of AB and CD. Thus, the greatest common measure of two given commensurable magnitudes, AB and CD, has been found. (Which is) the very thing it was required to show. H6pio\J.a. 'Ex 6r| xouxou cpavepov, oxi, edv [leye-Qoc, Suo jisys-dr] tiexprj, xal xo pieyiaxov auxwv xoivov ^expov [isxprpsi. Corollary So (it is) clear, from this, that if a magnitude measures two magnitudes then it will also measure their greatest common measure. 5'. Tpiwv ^leye'dcov au^iexpcov SoiSsvxwv xo [aeyioTov auxfiv xoivov jiexpov eupsTv. A' 1 B 1 r> 1 i — i i — i i — i A E Z TSaxw xd SoiJevxa xpia ^leye'dr] au^erxpa xd A, B, E- 8eT 8/) xfiv A, B, T xo [isyiaxov xoivov jisxpov supeiv. EiXrjcp'dto yap 8uo x«v A, B xo [isyiaxov xoivov ^xsxpov, xal screw xo A- xo 8/) A xo T fjxoi jiexpsl fj ou [jiexpsT]. ^expeixw itpoxepov. enel ouv xo A xo E [isxpei, [isxpeT Se Proposition 4 To find the greatest common measure of three given commensurable magnitudes. A i 1 B' ' C' 1 i 1 i — i i 1 D E F Let A, B, C be the three given commensurable mag- nitudes. So it is required to find the greatest common measure of A, B, C. For let the greatest common measure of the two (mag- nitudes) A and B have been taken [Prop. 10.3], and let it 285 ETOIXEIfiN i'. ELEMENTS BOOK 10 xal xd A, B, to A dpa xd A, B, T [lexpel/ xo A apa xwv A, B, r xoivov [iexpov eaxiv. xal cpavepov, oxi xal [icyiaxov [iclCov yap xou A [ieye , dou<; xa A, B ou [lexpel. Mr) [icxpeixo Sr] xo A xo r. Xey« xpGxov, oxi au[i[iexpd eaxi xa r, A. exel yap au[i[iexpd eaxi xa A, B, T, [iexpr]aei xi auxd [ieycdo<;, o Sr]XaSr) xal xa A, B [iexprjaei- &axe xal xo iwv A, B [icyiaxov xoivov [iexpov xo A [iexprjaei. [icxpeT Be xal xo E &axe xo eipr][ievov [icycdoc; [iexprjaei xa r, A- au[i[iexpa apa eaxl xa T, A. elXrjcp'dco ouv auxGv xo [icyiaxov xoivov [iexpov, xal eaxw xo E. exel ouv xo E xo A [iexpeT, dXXd xo A xd A, B [iexpeT, xal xo E dpa xa A, B [lexprpei. [iexpeT Se xal xo T. xo E dpa xd A, B, T [iexpeT/ xo E apa xwv A, B, T xoivov eaxi (iexpov. Xeyw oxi xal [icyiaxov. ei yap Buvaxov, eaxw xi xou E [iel£ov [leycdot; xo Z, xal [icxpeixw xd A, B, T. xal exel xo Z xd A, B, T [iexpeT, xal xd A, B apa (iexprjaei xal xo xov A, B [icyiaxov xoivov [iexpov [lexprpei. xo 8e xwv A, B (icyiaxov xoivov [iexpov eaxl xo A- xo Z apa xo A [iexpeT. [iexpeT Se xal xo r- xo Z apa xd T, A [iexpeT' xal xo xQv T, A apa [icyiaxov xoivov (iexpov [iexprjaei xo Z. eaxi 8e xo E- xo Z apa xo E [iexprjaei, xo [leT^ov xo eXaaaov oxep eaxlv dSuvaxov. oux apa [ieT£6v xi xou E [icye-doug [[leycdoc] xd A, B, T [iexpeT- xo E apa xwv A, B, T xo [icyiaxov xoivov [iexpov eaxlv, edv [ir] [icxpfj xo A xo T, edv 8e [icxpfj, auxo xo A. Tpiwv dpa [leyeiDwv au[i[iexp«v So'devxwv xo [icyiaxov xoivov [iexpov rjuprjxai [oxep e8ei Bel^ai]. be D. So D either measures, or [does] not [measure], C. Let it, first of all, measure (C). Therefore, since D mea- sures C, and it also measures A and B, D thus measures A, B, C. Thus, D is a common measure of A, B, C. And (it is) clear that (it is) also (the) greatest (common mea- sure). For no magnitude larger than D measures (both) A and B. So let D not measure C. I say, first, that C and D are commensurable. For if A, B, C are commensurable then some magnitude will measure them which will clearly also measure A and B. Hence, it will also measure D, the greatest common measure of A and B [Prop. 10.3 corr.]. And it also measures C. Hence, the aforementioned mag- nitude will measure (both) C and D. Thus, C and D are commensurable [Def. 10.1]. Therefore, let their greatest common measure have been taken [Prop. 10.3], and let it be E. Therefore, since E measures D, but D measures (both) A and B, E will thus also measure A and B. And it also measures C. Thus, E measures A, B, C. Thus, E is a common measure of A, B, C. So I say that (it is) also (the) greatest (common measure) . For, if possible, let F be some magnitude greater than E, and let it measure A, B, C. And since F measures A, B, C, it will thus also measure A and B, and will (thus) measure the greatest common measure of A and B [Prop. 10.3 corr.]. And D is the greatest common measure of A and B. Thus, F measures D. And it also measures C. Thus, F measures (both) C and D. Thus, F will also measure the greatest common measure of C and D [Prop. 10.3 corr.]. And it is E. Thus, F will measure E, the greater (measuring) the lesser. The very thing is impossible. Thus, some [magni- tude] greater than the magnitude E cannot measure A, B, C. Thus, if D does not measure C then E is the great- est common measure of A, B, C. And if it does measure (C) then D itself (is the greatest common measure). Thus, the greatest common measure of three given commensurable magnitudes has been found. [(Which is) the very thing it was required to show] H6pio\J.a. 'Ex 8r| xouxou (pavepov, oxi, edv [icye'doc; xpia (leye'dr) [lexpfj, xal xo [icyiaxov auxaiv xoivov [iexpov [iexprjaei. ! 0\±oiu>q Srj xal exl xXeiovwv xo [icyiaxov xoivov [iexpov Xrjcp-drjaexai, xal xo xopia[ia xpo)(iopr]aei. onep e8ei Sel^ai. Corollary So (it is) clear, from this, that if a magnitude measures three magnitudes then it will also measure their greatest common measure. So, similarly, the greatest common measure of more (magnitudes) can also be taken, and the (above) corol- lary will go forward. (Which is) the very thing it was required to show. 286 ETOIXEIftN i'. ELEMENTS BOOK 10 z. Proposition 5 Td atiuuexpa jieye'v}/] npbc, dXX/jXa Xoyov e/ei, ov Commensurable magnitudes have to one another the apiduoc; npbc, dpidfjiov. ratio which (some) number (has) to (some) number. A BT A B C i — i — i — i i — i — i i — i i — i — i — i i — i — i i — i "Eaxw auu^iexpa ^leye'dr] xa A, B- Xeyto, oxi to A Ttpoc to B Xoyov £X £l > ° v otpi'd^ioc; npbq dpid^iov. 'Enel yap ouujiSTpd eaxi Ta A, B, jiexprpei ti auxa [isys-Qoz. jiexpeixw, xal eaxio to T. xal oadxu; to T to A [lexpei, xoaauxai uovdBec eaxoaav ev xG A, oadxu; Be: to r to B [lexpel, xoaauxai uovdSec; eaxcoaav ev xa> E. Tkel ouv to T to A jiexpeT xaxd xac sv tw A uovd8a<;, ^texpeT Be xal f] uovac; xov A xaxd xa<; ev auxw ^ovdSac, ladxu; dpa f] ^tovac; tov A jiexpeT api'duov xal to T uey^ ? to A- eaxiv dpa w<; to T Ttpoc; to A, outck f] [Lovaz npbc, tov A- dvduaXiv dpa, cb? to A Tip6<; to T, oux«<; 6 A icpoc; t/]v ^.ovdBa. udXiv eirel to T to B ^expel xaxd xac; ev iu E ^lovdSac;, ^texpeT 8e xal f] uovdc; tov E xaTa xac; ev auxcp uovd8a<;, ladxu; dpa f] jxovac; tov E ^texpeT xal to T to B' eaTiv dpa w<; to T npbz to B, ouxmc; f) jiovdc; icpoc; tov E. eBelx'dr) 8e xal cbc; to A Tcpoc; to T, 6 A Tcpoc; t/]v jiovdBa- 8i° laou dpa eaTiv cbc; to A jcp6<; to B, ouxcoc; 6 A apiduoc; Ttp6<; tov E. Ta dpa au[i^expa ueye'dr) Ta A, B Tcpoc; aXXrjXa Xoyov exei, Sv dpnf)[i6<; 6 A icpoc dpi%6v tov E- oicep eBei SeT^ai. Let A and B be commensurable magnitudes. I say that A has to B the ratio which (some) number (has) to (some) number. For if A and B are commensurable (magnitudes) then some magnitude will measure them. Let it (so) measure (them), and let it be C. And as many times as C measures A, so many units let there be in D. And as many times as C measures B, so many units let there be in E. Therefore, since C measures A according to the units in D, and a unit also measures D according to the units in it, a unit thus measures the number D as many times as the magnitude C (measures) A. Thus, as C is to A, so a unit (is) to D [Dei. 7. 20] J Thus, inversely, as A (is) to C, so D (is) to a unit [Prop. 5.7 corr.]. Again, since C measures B according to the units in E, and a unit also measures E according to the units in it, a unit thus measures E the same number of times that C (measures) B. Thus, as C is to B, so a unit (is) to E [Def. 7.20]. And it was also shown that as A (is) to C, so D (is) to a unit. Thus, via equality, as A is to B, so the number D (is) to the (number) E [Prop. 5.22]. Thus, the commensurable magnitudes A and B have to one another the ratio which the number D (has) to the number E. (Which is) the very thing it was required to show. t There is a slight logical gap here, since Def. 7.20 applies to four numbers, rather than two number and two magnitudes. Edv 860 [leye'f)/) Ttpoc; dXXrjXa Xoyov eyjr), ov dpidjjidc; Ttpoc; dpi%6v, au^expa eaxai Ta [leye'dr]. A 1 — 1 — 1 — 1 — 1 B 1 1 A' ' E' 1 pi — 1 Z' — 1 — 1 — 1 Auo yap fieye'dr] xd A, B Ttpoc; dXXr)Xa Xoyov e^exco, ov dpiOjjiog 6 A Ttpoc; dprd^iov tov E- Xeyw, 6x1 atiunexpa eaxi Ta A, B ^eyei}/]. "Oaai yap eicfiv ev tu A ^tovdSec;, ek ToaaOTa laa Proposition 6 If two magnitudes have to one another the ratio which (some) number (has) to (some) number then the magni- tudes will be commensurable. A I — I — I — I 1 B 1 1 D' 1 E i 1 C 1 — 1 F 1 — i — i — 1 For let the two magnitudes A and B have to one an- other the ratio which the number D (has) to the number E. I say that the magnitudes A and B are commensu- rable. 287 ETOIXEIfiN i'. ELEMENTS BOOK 10 SirpiqcrdM to A, xal evi auTGv Taov sgtw to r - oaai 6e rimv sv to E [iovd8ec, ex toooutmv [icycdGv lacov to T auyxeiai&M to Z. 'EtieI ouv, oaai eioiv ev tG A (iovd8ec, ToaauTa rim xai ev tG A ^eyedr) ^ aa T Q r, o dpa jiepoc taxiv f] jiovac tou A, to auTo [izpoc, ecru xal to T tou A- eaTiv apa Gc to T Tipoc to A, outgjc f) jiovac Tipoc tov A. 8s f) (iovac tov A dpiOjiov (iETpsI dpa xal to V to A. xal ene'i eaTiv Gc to T Tipoc to A, outioc f) ^ovdc Ttpoc tov A [dpidjiov], dvaTiaXiv apa Gc to A Ttpoc to T, outwc 6 A dpidjioc Ttpoc tt)v [iovd8a. TtdXiv insi, oaai eialv ev ifi E [iovdSec, ToaauTa eloi xal ev tG Z I'oa tG T, eaTiv apa Gc to T Tipoc to Z, outcoc f) ^xovac Ttpoc tov E [dpi%6v]. eSei/i}/) 8e xal Gc to A Ttpoc to r, outcoc 6 A Ttpoc tt)v [iovd5a' 5i' laou apa cotIv Gc to A Ttpoc to Z, outoc 6 A Ttpoc tov E. dXX'' Gc 6 A Ttpoc tov E, outw<; cotI to A Ttpoc to B - xal Gc dpa to A Ttpoc to B, outwc xal Ttpoc to Z. to A apa Ttpoc exaTcpov tGv B, Z tov ai)Tov §)(ei Xoyov I'aov apa eaTi to B tG Z. [iCTpel 8e to T to Z - ^.eTpeT dpa xal to B. dXXd \ir\M xal to A' to r dpa Ta A, B [iCTpeT. aujijiCTpov dpa sotI to A tG B. 'Eav dpa 8uo (jieyeiSr) Tipoc dXXrjXa, xal Ta ec"rjc- LTopiajJia. 'Ex 8r) toutou cpavepov, oti, eav Gai 8uo dpi , d|ioi, Gc oi A, E, xal suiDeTa, Gc f) A, SuvaTov eaTi Tioifjaai Gc 6 A dpid^ioc Ttpoc tov E dpi/d^iov, outoc ttjv eu-delav Tipoc euif)eTav. eav Se xal tGv A, Z [Leaf] dvdXoyov Xrjcp'dfj, Gc rj B, soTai Gc f) A Ttpoc; t?]v Z, outoc to dito xrjc; A Ttpoc; to aTco Tfjc B, touteotiv Gc f) TcpGTT) Ttpoc t/]v Tp[T/]v, outoc to aTco Tfjc TtpGTTjc Ttpoc; to aTto Tfjc BeuTcpac to ojioiov xal o^toiwc dvaypacpo^tevov. dXX' Gc; f\ A Ttpoc t/]v Z, outok eaTiv 6 A dpid^oc; upoc; tov E dpn!)^6v ysyovev dpa xal Gc; 6 A dpiiD^toc; Tipoc; tov E dpnD^tov, outwc; to dito Tfjc A eu-delac Ttpoc to duo Tfjc B eu^eiac oTcep e8ei 8ric;ai. Ta douji|jieTpa ^ieyei9r) Tipoc dXXr)Xa Xoyov oux ex ei J ° v dpn!)^6c Tipoc dpidjiov. 'EaTW dau^i^ieTpa ^.eye'dr] Ta A, B- Xeyco, oti to A Tipoc to B Xoyov oux exei, 8v dpi-djioc Tipoc dpid^tov. For, as many units as there are in D, let A have been divided into so many equal (divisions). And let C be equal to one of them. And as many units as there are in E, let F be the sum of so many magnitudes equal to C. Therefore, since as many units as there are in D, so many magnitudes equal to C are also in A, therefore whichever part a unit is of D, C is also the same part of A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. And a unit measures the number D. Thus, C also measures A. And since as C is to A, so a unit (is) to the [number] D, thus, inversely, as A (is) to C, so the number D (is) to a unit [Prop. 5.7 corr.]. Again, since as many units as there are in E, so many (magnitudes) equal to C are also in F, thus as C is to F, so a unit (is) to the [number] E [Def. 7.20]. And it was also shown that as A (is) to C, so D (is) to a unit. Thus, via equality, as A is to F, so D (is) to E [Prop. 5.22]. But, as D (is) to E, so A is to B. And thus as A (is) to B, so (it) also is to F [Prop. 5.11]. Thus, A has the same ratio to each of B and F. Thus, i? is equal to F [Prop. 5.9]. And C measures F. Thus, it also measures B. But, in fact, (it) also (measures) A. Thus, C measures (both) A and B. Thus, ^4 is commensurable withB [Def. 10.1]. Thus, if two magnitudes ... to one another, and so on Corollary So it is clear, from this, that if there are two numbers, like D and E, and a straight-line, like A, then it is possible to contrive that as the number D (is) to the number E, so the straight-line (is) to (another) straight-line (i.e., F). And if the mean proportion, (say) B, is taken of A and F, then as A is to F, so the (square) on A (will be) to the (square) on B. That is to say, as the first (is) to the third, so the (figure) on the first (is) to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.]. But, as A (is) to F, so the number D is to the number E. Thus, it has also been contrived that as the number D (is) to the number E, so the (figure) on the straight-line A (is) to the (similar figure) on the straight-line B. (Which is) the very thing it was required to show. Proposition 7 Incommensurable magnitudes do not have to one an- other the ratio which (some) number (has) to (some) number. Let A and B be incommensurable magnitudes. I say that A does not have to B the ratio which (some) number (has) to (some) number. 288 ETOIXEIftN i'. ELEMENTS BOOK 10 A' ' B 1 El yap zyzi xo A npo<; to B Xoyov, ov dtpi , d(Ji6c; upoc; dpi%6v, auji^texpov eaxai xo A iw B. oux eoxi Be- oux apa xo A 7ip6<; xo B Xoyov zyzi, ov apiduoc; npoz dpid^ov. Td apa dau^[iexpa ^teys'dr] 7ipo<; aXXrjXa Xoyov oux e/ei, xal xd eZ>y]z- *)'• 'Edv 8uo (ieyeiSr) npbc, dXX/]Xa Xoyov [ir\ z^f], 8v dpi'd^oc; npbc, apiduov, dau^iexpa eaxai xa ^eyei&r]. A' ' B 1 Auo yap ^xsyei}/) xa A, B Tipoc; aXXrjXa Xoyov [ir] E)(£x(o, ov dpi'djj.ot; upog apidjjiov Xeyw, oxi dau^expd eaxi xa A, B ^sysdr). Ei yap Eaxai aujijiexpa, xo A upog xo B Xoyov zZ